A diver leaves the end of a 5.00 m high diving board and reaches a maximum height of 5.50 m before she strikes the water below. The landing point is 3.25 m beyond the end of the diving board in the horizontal direction. You may approximate the diver as a point particle. Hint: Carefully consider your placement of the origin before starting your calculations.
Determine:_____.
(a) After the diver leaves the diving board, how long does it take for her to strike the water?
(b) The initial velocity of the projectile 7. (hint: a vector has both direction and magnitude).
(c) The final velocity with which the diver enters the water ö, (hint: a vector has both direction and magnitude).

Answers

Answer 1

Answer:

a) 1.38 s

b) v₀ = 3.92 m/s  θ₀ = 53.1º

c) vf = 10.7 m/s θf = -77.2º

Explanation:

a)

Once the diver leaves the board, she has an initial velocity vector, that we can split in two directions perpendicular each other, the vertical direction (coincident with the y-axis, positive upward) and the horizontal direction (coincident with the x-axis, positive away from the diving board).These velocities, are independent each other, due to they are perpendicular.In the horizontal direction, since no acceleration takes place, she keeps the horizontal component v₀ₓ, constant.In the vertical direction, she is always accelerated downward by gravity, which constant value is g = 9.8m/s2.With this all information, we can find the total time since she left the diving board till she strikes the water below, taking advantage of the fact that she reached to a maximum height of 5.5 m,i.e. , 0.5m above the diving board. When at this point, the vertical component of the velocity is zero, so we can write the following kinematic equation:

       [tex]v_{fy} ^{2} - v_{oy} ^{2} = 2* g* \Delta y (1)[/tex]

Solving for v₀y in (1):

       [tex]v_{oy} = \sqrt{2*g*\Delta_{ymax} } = \sqrt{2*9.8m/s2*0.5m} = 3.13 m/s (2)[/tex]

When she is at the highest point, her final vertical speed is zero, so we can find the time when she reached to this point, applying the definition of acceleration, as follows:

       [tex]v_{fyhmax} = v_{oy} - g*t_{hmax} = 0 (3)[/tex]

Solving for t in (3), we get:

       [tex]t_{hmax} =- \frac{v_{oy}}{g} =\frac{3.13m/s}{9.8m/s2} = 0.32 s (4)[/tex]

Since at this point, the vertical speed is zero, we can apply the equation for the displacement (vertical in this case) for a total displacement of 5.5 m (taking as the origin the point just below the end of the diving board, on the surface of the water), with initial speed equal to zero, as follows:

       [tex]\Delta y = \frac{1}{2}* g * t^{2} = 5.5 m (5)[/tex]

Solving for t (which we will call t₂) in (5), we get:

       [tex]t_{2} = \sqrt{\frac{2*\Delta y}{g}} = \sqrt{\frac{2*5.5m}{9.8m/s2} } = 1.06 s (6)[/tex]

The total time will be just the sum of (4) (since leaving the diving board till reaching the maximum height) and (6) (from there until she strikes the water), as follows:Δt = thmax + t₂ = 0.32 s + 1.06 s = 1.38 s (7)

b)

In order to know the value of v₀, we need first to get the values of vox and voy.We know that the horizontal speed remains constant, so, if we know the total time traveled (same as in a)) and the horizontal distance (given), we can find v₀ₓ applying the definition of average velocity, as follows:

      [tex]v_{ox} =\frac{\Delta x}{\Delta t} =\frac{3.25m}{1.38s} = 2.36 m/s (8)[/tex]

We know the value of v₀y from (2):

       [tex]v_{oy} = 3.13 m/s (9)[/tex]

Applying the Pythagorean theorem, we can get the magnitude of v₀ as follows:

      [tex]v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{(2.36m/s)^{2} + (3.13m/s)^{2}} = 3.92m/s (10)[/tex]

We can find the angle that it makes with the horizontal, applying the definition of tangent of an angle, as follows:

       [tex]tg_{(\theta o)} = \frac{v_{oy}}{v_{ox} } = \frac{3.13}{2.36} = 1.33 (11)[/tex]

       ⇒ θ₀ = tg⁻¹ (1.33) = 53.1º

c)

In order to find the vertical component of the final velocity, we can use the following kinematic equation, for the total vertical displacement of 5.5 m since it started to fall after reaching to the maximum speed:

        [tex]v_{fy} ^{2} - v_{oy} ^{2} = 2* g* \Delta y (12)[/tex]

When it started to fall, in the vertical direction, the diver was at rest, so in (12) we have v₀y = 0.Solving for vfy in (12) we get:

       [tex]v_{fy} = \sqrt{2*g*\Delta_{y} } = \sqrt{2*9.8m/s2*5.5m} = -10.4 m/s (13)[/tex]

We already know that the horizontal component is the same as the initial v₀ₓ, so we can find the magnitude of the final velocity applying the Pythagorean Theorem as follows:

       [tex]v_{f} = \sqrt{v_{fx}^{2} + v_{fy}^{2}} = \sqrt{(2.36m/s)^{2} + (-10.4m/s)^{2}} = 10.7 m/s (14)[/tex]

Finally, as we did in (11) to find the angle that it makes with the horizontal, we can find θf as follows:

       [tex]tg_{(\theta f)} = \frac{v_{fy}}{v_{fx} } = \frac{-10.4}{2.36} = -4.41 (15)[/tex]

      ⇒ θf = tg⁻¹ (-4.41) = -77.2º


Related Questions

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Answer:

We answer that before but I forgot

You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height h from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 m/s when they reach the bottom of the ramp. You determine that for a 77.0 kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope.

Required:
What is the maximum height h for which the maximum safe speed will not be exceeded?

Answers

Answer:

h = 40.37 m

Explanation:

We will apply the law of conservation of energy to the skier in this case, as follows:

[tex]Energy\ of\ skier\ at\ the\ gate = Energy\ of\ Skier\ at\ the\ end\\P.E + K.E_{i} = K.E_{f} - W_{friction}\\mgh + \frac{1}{2}mv_{i}^2 = \frac{1}{2}mv_{f}^2 - W_{friction}\\\\mgh = \frac{1}{2}m(v_{f}^2-v_{i}^2) - W_{friction}[/tex]

where,

m = mass of skier = 77 kg

g = acceleration due to gravity = 9.81 m/s²

vf = final speed = 30 m/s

vi = initial speed = 2 m/s

W_friction = Work done by friction and air resistance = 4000 J

Therefore,

[tex](77\ kg)(9.81\ m/s^2)h = \frac{1}{2}(77\ kg)[(30\ m/s)^2-(2\ m/s)^2] - 4000\ J\\\\h = \frac{34496\ J - 4000\ J}{755.37\ N}\\\\[/tex]

h = 40.37 m

Method
1. Place a trolley on a smooth runway and tilt the runway very slightly so that the
slope compensate for the friction between the wheels and the runway. If
pushed gently, the trolley should move at a constant velocity.
2. Measure the mass of the trolleys.
3. Attach the ticker timer to the trolley and thread the tape through the ticker
timer.
4. Attach a smooth pulley vertically to the bottom of the runway.

Answers

Answer:

3

Explanation:

it's too because on you measuring mass of something

identify and explain the climate cycle shown
below. Illustrate how conditions change during
the summer

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Answer:

wnd change

Explanation: wnd chsnge

Which statement about acceleration is FALSE?
Group of answer choices

A. A car which is slowing down is accelerating

B. When acceleration and velocity are in opposite directions, acceleration is positive

C. If an object changes its direction of motion, its acceleration changes

D. A car which is speeding up is accelerating

Answers

answer: A) a car which is slowing down is accelerating

that is a false statement

8. Two perpendicular forces, one of 45.0 N directed upward and the other of 60.0 N directed to the
right, act simultaneously on an object with a mass of 35.0 kg. What is the magnitude of the
resultant acceleration of the object?
a) 2.14 m/s
b) 5.25 m/s
c) 3.00 m/s
d) 1.41 m/s

Answers

Answer:

[tex]a=2.14\ m/s^2[/tex]

Explanation:

Force in upward direction = 45 N

Force in right side = 60 N

Mass of the object, m = 35 kg

Net force on the object is given by :

[tex]F=\sqrt{45^2+60^2} \\F=75\ N[/tex]

We know that,

F = ma

Where

a is acceleration of the object

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{75\ N}{35\ kg}\\\\a=2.14\ m/s^2[/tex]

So, the acceleration of the object is [tex]2.14\ m/s^2[/tex].

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Answers

A, and D are two correct answers im not sure about the third sorry

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Answers

Answer:

14 m/s^2

Explanation:

We have to use the equation F=ma because the question is a Newton's second law question.

First, list known values

F (force) = 70 N

m (mass) = 5 kg

a (acceleration) = ? (we have to solve for this)

Second, plug in known values into equation.

F=ma

70=5a

Third, solve. To solve this all you have to do is divide both sides by 5

70/5=5/5a

Fourth, your answer will be 14 meters per second squared  (m/s^2)

I hope this answer helps and have a great day!

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7.5 m/s just multiply the two values

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True or False

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Answer:

True

Newton's Rainbow. In the 1660s, English physicist and mathematician Isaac Newton began a series of experiments with sunlight and prisms. He demonstrated that clear white light was composed of seven visible colors.

Explanation:

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Answer:

Plasma is a phase of matter that makes up stars.

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ωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωωZ?

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Answers

Answer:524 Hz

Explanation:

Approximate frequency, heard in other car, when two car approaches each other, before they cross each other is 524 Hz.

What is frequency?

Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time.

For the two cars approaching each other the Doppler formula to find the frequency of second car is given as,

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Here, [tex]V_s[/tex] is the speed of the sound.

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Convert the unit of velocity of the car as,

[tex]\rm 54km/s=54\times\dfrac{5}{18}m/s\\\rm 54km/s=15m/s[/tex]

As we know that the speed of the sound is 340 m/s and one of the cars emits a note of frequency 480 Hz.

Thus, putting the values in the above formula to find out the frequency heard in the other car before they cross each other as,

[tex]f_2=\dfrac{340+15}{340-15}480\\f_2=524.3\rm Hz[/tex]


The approximate frequency heard in the other car before they cross each other is 524 Hz.

Learn more about the frequency here;

https://brainly.com/question/1436295

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The answer is Non contract force

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Answers

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Answers

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f(3) = 362.32 Hz

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0 for the first

1 for the second

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Answers

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forecasted for the miscommunication between range and the other day

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b cuz i did the that

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directly proportional

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Answers

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Answers

Answer:

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