Answer:
a) the weight of the person is 170.5 lbf
b) weight of the astronaut on the moon is 28.2 lbf
Explanation:
Given the data in the question;
a)
we know that;
weight on the surface of the earth = m[tex]g_{earth[/tex]
given that m = 170.5 lbm and g = 32.174 ft/s²
we substitute
weight on the surface of the earth = 170.5 lbm × 32.174 ft/s²
= 5485.667 lbm-ft/s²
1 lbf = 32.174 lbm-ft/s²
so
weight on the surface of the earth = (5485.667 / 32.174) lbf
weight on the surface of the earth = 170.5 lbf
Therefore, the weight of the person is 170.5 lbf
b)
given that;
weight on the surface of the earth = m[tex]g_{moon[/tex]
m = 170.5 lbm and g = 5.32 ft/s²
weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²
= 907.06 lbm-ft/s²
1 lbf = 32.174 lbm.ft/s²
weight on the surface of the earth = ( 907.06 / 32.174 ) lbf
weight on the surface of the earth = 28.2 lbf
Therefore, weight of the astronaut on the moon is 28.2 lbf
Explain what happens to the movement, energy and arrangement of particles during the process
of distillation.
Answer:
Distillation separates a liquid from a solution. For example, water can be separated from salty water by simple distillation. This method works because the water evaporates from the solution, but is then cooled and condensed into a separate container. The salt does not evaporate and so it stays behind.
During which sequence of phases does the moon's visibility increase
During the "waxing" phases, the two weeks immediately following the New Moon.
New Moon ==> waxing crescent ==> First Quarter ==> waxing gibbous ==> Full Moon
The phase of the moon decides the moon's visibility. The sequence of phases does the moon's visibility increases is; the waxing crescent, first quarter, waxing gibbous, full Moon.
What is the moon phase?The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.
Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.
We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.
Following are the sequence of phases does the moon's visibility increase is;
1. Waxing crescent
2. First Quarter
3. Waxing gibbous
4. Full Moon
The phase of the moon decides the moon's visibility. Hence the visibility changes with the change in the phase of the moon.
To learn more about the moon's phase refer to the link;
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What x rays travel at the speed of
The sun warms Earth through the process of _________
conduction
convection
insulation
radiation
14. Which of the following is not an example of work being done?
A. pushing a basketball away from your body
B. holding a coffee mug
C. carrying boxes across a warehouse floor
Answer:
B. holding a coffee mug
Explanation:
Something must move a distance for work to be done.
A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.
Answer:
The acceleration is 2.448 meters per square second and is vertically upward.
Explanation:
The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:
[tex]\Sigma F = F - m\cdot g = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Buoyant force, measured in newtons.
[tex]m[/tex] - Mass of the plastic ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration, measured in meters per square second.
If we know that [tex]F = 5\,N[/tex], [tex]m = 0.408\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the net acceleration of the plastic ball is:
[tex]a = \frac{F}{m} - g[/tex]
[tex]a= 2.448\,\frac{m}{s^{2}}[/tex]
The acceleration is 2.448 meters per square second and is vertically upward.
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
In what order were the following energy sources
discovered by humans
In which scenario is the greatest amount of work done on a wagon?
A force of 55 N moves it 8 m.
A force of 60 N moves it 6 m.
A force of 50 N moves it 5 m.
A force of 40 N moves it 10 m.
Answer:
The first scenario!
Explanation:
W=F*d
a) 55*8= 440J
b) 60*6= 360J
c) 50*5= 250J
d) 40*10= 400J
The scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.
What is Work done?Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
W = F × d
Where F is force applied and d is distance travelled.
From the question;
A force of 55 N moves it 8 m
W = 55N×8m = 440N.m
A force of 60 N moves it 6 m.
W = 60N×6m = 360N.m
A force of 50 N moves it 5 m.
W = 50N×5m = 250N
A force of 40 N moves it 10 m.
W = 40N×10m = 400N.m
Therefore, the scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.
Learn more about work done: brainly.com/question/26115962
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is electricity matter
Answer:
Yes it is
Explanation:
Electricity is the positive and negative matter that's found in protons and electrons.
Answer:
yes
Explanation:
because electricity is a positive and negative proton
Please answer :>
40 POINTS
Answer:
rotates
Explanation:
I'm so bored
yrfgggghhgghhyuj
coefficient of viscosity of a glycerine is 8.4 poison explain
Answer:
coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.
Explanation:
Viscosity is defined as the extent to which a fluid can resist flow when a force is applied to it.
Now, coefficient of viscosity is the term in which viscosity is calculated. It is basically the tangential frictional force acting per unit area which is divided by the velocity gradient as a result of streamline flow conditions.
Thus, coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.
PLEASE HELP!!!
What are some compound machines that are used today?
Answer:
Examples of compound machines include bicycles, cars, scissors, and fishing rods with reels. Compound machines generally have lower efficiency but greater mechanical advantage than simple machines
Brilianst
Answer:
screws, inclined planes , etc
Explanation:
no explanation needed
(What is the weight of a 50 kg woman at the top of Jupiter's
atmosphere, where g = 24.8 N/kg? Give your answer in both
newtons and pounds.
Answer:
The correct answer is - 1240 newtons; 278.763 lbs.
Explanation:
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
thank you
Answer:
Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.
Answer:
Hey mate...
Explanation:
This is ur answer....
In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.
Hope it helps you,
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3. A woman drove her car from home to her daughter's school. The odometre on her dashboard says she travelled 4.5 km to do this. She then immediately drove back home, using a different route, which was 5.5 km long. The whole journey took 30 minutes.
a. What distance did she travel?
b. What was her displacement?
C. What was her average speed during the journey?
Answer:
Look Below -->
Explanation:
a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )
b. Her displacement is 0 km because she went back home. (Displacement is the difference between the end and starting points)
c. 3 km/hr (30 minutes / 10 km)
Which of the following requires constant agonist-antagonist muscle contraction
Answer: Dynamic balance
Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121
What happens when a moving object experiences no net force?
Answer:
An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.
The potential energy of a 40-kg cannonball is 14000 J. How high was the cannon ball to have this much potential energy?
Answer:
35.71 m
Explanation:
Potential energy is calculated using this formula:
PE = mghwhere m = mass (kg). g = gravitational acceleration on Earth (9.8 m/s²), h = height (m)We are given 3 out of the 4 variables in this problem. We want to solve for h, the height of the cannon ball.
List out the known variables:
PE = 14,000 J m = 40 kg g = 9.8 m/s² h = ? mSubstitute these values into the potential energy formula.
14,000 = (40)(9.8)h 14,000 = -392h h = 35.7142857143The cannonball was 35.71 m high to have a potential energy of 14,000 J.
g We have studied diffraction from a single slit, where light is sent through a thin opening. A similar phenomena occurs when light bends around a thin object, like a human hair. Here the width of the hair plays the role of the width of the single slit. Measurements found that when a beam of light of wavelength 632.8 nm was shone on a single strand of hair, the first dark fringe on either side of the central bright spot were 5.22 cm apart. If the screen is 1.25 meters away, how thick was this strand of hair?
Answer:
[tex]3.031\times 10^{-5}\ \text{m}[/tex]
Explanation:
[tex]y[/tex] = Distance between central maxima and first minimum
m = Order = 1
d = Thickness of hair
[tex]\lambda[/tex] = Wavelength = 632.8 nm
L = Distance between light source and screen = 1.25 m
Width of central maximum is given by
[tex]2y=5.22\times 10^{-2}\\\Rightarrow y=\dfrac{5.22\times 10^{-2}}{2}\\\Rightarrow y=0.0261\ \text{m}[/tex]
Distance between central maxima and first minimum is given by
[tex]y=L\tan\theta_{min}\\\Rightarrow \tan\theta_{min}=\dfrac{y}{L}\\\Rightarrow \tan\theta_{min}=\dfrac{0.0261}{1.25}\\\Rightarrow \theta_{min}=\tan^{-1}0.02088\\\Rightarrow \theta_{min}=1.1962^{\circ}[/tex]
Since [tex]\theta[/tex] is small [tex]\tan\theta_{min}=\sin\theta_{min}[/tex]
[tex]\sin\theta_{min}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1\times 632.8\times 10^{-9}}{\sin1.1962^{\circ}}\\\Rightarrow d=3.031\times 10^{-5}\ \text{m}[/tex]
The strand of hair is [tex]3.031\times 10^{-5}\ \text{m}[/tex] thick.
Plsss I want answer???
Answer:
s=136.89/2g meter
s=6.98 meter (correct to 3 sig.fig. taking g=9.81ms^-2)
Explanation:
u= + 11.7 ms^-1
a= - g ms^-2
At highest point: v=0ms^-1
v^2=u^2+2as
0=11.7^2+2(-g)s
s=136.89/2g meters
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of m A = 17.0 kg and an initial velocity of v 0A = 8.00 m/s, due east. Object B, however has a mass of m B = 29.0 kg and an initial velocity of v 0B = 5.00 m/s, due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.
Answer:
pf = 198.8 kg*m/s
θ = 46.8º N of E.
Explanation:
Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:[tex]p_{ox} = p_{oAx} + p_{oBx} (1)[/tex]
We can do exactly the same for the initial momentum along the y-axis:[tex]p_{oy} = p_{oAy} + p_{oBy} (2)[/tex]
The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:[tex]p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)[/tex]
We can repeat the process for the y-axis, as follows:[tex]p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)[/tex]
Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:[tex]v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)[/tex]
In the same way, we can find the component of the final momentum along the y-axis, as follows:[tex]v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)[/tex]
With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)[/tex]
The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:[tex]p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)[/tex]
Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.We can find this angle applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)[/tex]
⇒ θ = tg⁻¹ (1.06) = 46.8º N of E
You are riding in the passenger seat of a car as it goes around a tight turn. You slide across the seat to the passenger side door. Which statement below properly describes what is happening? a. You are exerting a centripetal force on the door. b. The door and seat are exerting a centripetal force on you that balances the centrifugal force of the turn. c. The car seat exerts a centripetal force on you, but not enough to keep you in place, so the door exerts the rest. d. The centrifugal force is greater than the force of friction between you and the seat, so you slide outward.
Answer:
Explanation:
answer C looks good
there isn't really a "centrifugal " force. :/ when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too, a force pushes us into the corner, not a force pushing out of the corner. :)
An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is downward. towards the west. upward. towards the south. towards the north.
Answer:
towards the north.
Explanation:
The uniform magnetic field has strength and direction in all points. The upwards motion of the field or he electronic curves will show a northern direction. Hence, the field can be created by taking the opposite magnetic in the two directions. The straight line of motion of the field will indicate the flow in the north direction. The magnetic field lines will remain parallel and stay uniform to poles.A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.
Answer:
a. 3.73 m/s b. 27.8 m/s²
Explanation:
(a) Calculate his velocity (in m/s) when he leaves the floor.
Using the conservation of energy principles,
Potential energy gained by basketball player = kinetic energy loss of basket ball player
So, ΔU + ΔK = 0
ΔU = -ΔK
mg(h' - h) = -1/2m(v'² - v²)
g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor
Substituting the values of the variables into the equation, we have
9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)
9.8 m/s²(0.710 m) = -1/2(-v²)
6.958 m²/s² = v²/2
v² = 2 × 6.958 m²/s²
v² = 13.916 m²/s²
v = √(13.916 m²/s²)
v = 3.73 m/s
(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.
Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m
So, making, a subject of the formula, we have
a = (v² - u²)/2s
Substituting the values of the variables into the equation, we have
a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)
a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)
a = 13.913 m²/s²/(0.50 m)
a = 27.83 m/s²
a ≅ 27.8 m/s²
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
A friend pushes a sled across horizontal snow and when it gets up to speed the friend jumps on. After the friend jumps on, the sled gradually slows down. Which forces act on the combined sled plus friend after the friend jumps on
Answer:
v’ =( [tex]\frac{1}{1+ \frac{M}{m} }[/tex] ) v
we see that the greater the difference, the more the sled slows down.
friction force
Explanation:
When the man pushes the sled he does work and the sled acquires a speed and as long as it is supplied with an energy equal to the work of the chipping force with the snow, the speed is maintained.
When he jumps on the sled, a collision occurs and the initial moment
p₀ = mv
is increased by the increase in mass
m_f= (m + M_{man} ) v '
In this case there is no longer any external force applied and the only external force is friction, which causes the sled to stop, even when it is small, but the significant reduction in speed is due to the increase in masses.
p₀ = p_f
mv = (m + M_{man}) v '
v ’= [tex]\frac{m}{m+M}[/tex] v
v’ =( [tex]\frac{1}{1+ \frac{M}{m} }[/tex] ) v
Therefore, we see that the greater the difference, the more the sled slows down.
The only forces that act on the sled with the man are the friction that is responsible for the decrease in speed and weight with the normal
What happens when a moving object is moving a particular direction experiences a net force opposite the direction?
Answer:
stop
Explanation:
object stop
Answer:
If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement. Two or more opposite forces are balanced forces if their effects cancel each other and they do not cause a change in an object's motion.
What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
m/s2?
Answer:Mass of the body = 20 kg.
Final Velocity = 5.8 m/s.
Initial velocity = 0
Time = 3 seconds.
Using the Formula,
Acceleration = (v - u)/ t
= (5.8 - 0)/ 3
= 1.6 m /s².
Now, Using the Formula,
Force = mass × acceleration
= 20 × 1.6
=
Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]
The magnitude of requires force, that is exerted on a 20 kg mass to give it an acceleration of 10.0 m/s^2 is 200 Newton.
What is force?The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.
Given parameters:
Mass of the object: m = 20 kg.
Acceleration of the object: a = 10.0 m/s^2.
Hence, according to Newton's 2nd law of motion:
the magnitude of requires force = mass ×acceleration
= 20 × 10 Newton
= 200 Newton.
Hence, the magnitude of requires force is 200 Newton.
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Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.