Gravitational potential energy is greatest at the highest point of a roller coaster and least at the lowest point.

Answer:

 θ₁ = 0.04º , θ₂ = 0.00118º

Explanation:

The equation that describes the diffraction pattern of a network is

             d sin θ = m λ

where the diffraction order is, in this case they indicate that the order

m = 1

           θ = sin⁻¹ (λ / d)

Trfuvsmod ls inrsd fr ll red s SI units

           d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm

the distance between two lines we can look for it with a direct proportions rule

If there are 4724 lines in a centimeter, the distance for two hundred is

            d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm

let's calculate the angles

λ = 300 10-9 m

            θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)

            θ₁ = sin⁻¹ (7.08 10-4)

            θ₁ = 0.04º

λ = 5000

          θ₂ = sin-1 (500 10-9 / 4,2337 10-4)

          θ₂ = 0.00118º

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

c.they have opposite charges.

Explanation:

because the protons have a positive charge and the neutrons have no charge.

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

Answer:

The gray spheres is negatively charged while the red is positively charged

Explanation:

This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged

Answer:

The gray sphere has a positive charge and the red sphere has a positive charge.

Answer:

The two masses are 3.39 Kg and 1.75 Kg

Explanation:

The gravitational force of attraction between two bodies is given by the formula;

F = Gm₁m₂/d²

where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²

m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects

Further calculations are provided in the attachment below

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Answer:

A: An upward force balances the downward force of gravity on the skydiver.

Explanation:

on edge! hope this helps!!~ (⌒▽⌒)☆

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

[tex]\theta=90-40=50^o[/tex]

From the triangle OAB

[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]

[tex]1340.57=35^2+25^2-x^2[/tex]

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

Answer:

In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Explanation:

In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.

Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Answer:

a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]

b) Q = 45 µC

c) C' = 1.5 μF

d)  [tex]E = 6.75 *10^{-4} J[/tex]

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]

Answer:

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = [tex]\frac{I}{nqA}[/tex]

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = [tex]\frac{I}{nqA}[/tex]

v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

Answer:

 θ  = 180

Explanation:

When an electric dipole is placed in an electric field, there is a torque due to the electric force

           τ = p x E

by rotating the dipole there is a change in potential energy

        ΔU = ∫ τ dθ

        ΔU = p E (cos θ₂ - cos θ₁)

when the dipole starts from an angle to the equilibrium position for θ = 0

          ΔU = pE (cos θ  - cos 0)

           cos θ  = 1 + DU / pE)

let's apply this expression to our case, the change in potential energy is ΔU = -0.2J

let's calculate

          cos θ  = 1 -0.2 / 0.1

          cos θ  = -1

           θ  = 180

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]

where

[tex]S_A[/tex] indicates the speed of the ship A

[tex]S_B[/tex] indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where K indicates spring constant

m indicates mass

For the new time period

[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]

Now, we will take 2 ratios of the time period

[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]

[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]m' = \frac{0.500}{0.5625}[/tex]

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.

Given :

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.

The formula of the time period is given by:

[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex]   ---- (1)

where m is the mass and K is the spring constant.

The new time period is given by:

[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex]   ---- (2)

where m' is the total mass after the addition and K is the spring constant.

Now, divide equation (1) by equation (2).

[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]

Now, substitute the known terms in the above expression.

[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]

Simplify the above expression in order to determine the value of m'.

[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]

m' = 0.889 Kg

Now, the mass added to the object to change the period to 2.00 s is given by:

m" = 0.889 - 0.500

m" = 0.389 Kg

For more information, refer to the link given below:

https://brainly.com/question/2144584

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

Answer:

Since the angular acceleration of the objects will be proportional to the torque (due to gravity) acting on them and they will all experience the same torque their accelerations will be inversely proportional to their moments of inertia:

I disk = 1/2 M R^2

I sphere = 2/5 M R^2

I shell = 2/3 M R^2

Thus the sphere will experience the greatest angular acceleration and reach the bottom first, and then be followed by the disk and the shell.

By conservation of energy they will all have the same kinetic energy when they reach the bottom of the ramp.

(a) The ranking of the objects in order of how they will finish the race is

solid sphere > disk > hollow spherical shell

(b) The ranking of the objects in order of kinetic energy is

solid sphere > disk > hollow spherical shell

The moment of inertia of each object is calculated as follows;

The angular momentum of the objects is calculated as follows;

[tex]L =I \omega \\\\\omega = \frac{L}{I}[/tex]

The object with the least moment of inertia is will have the highest speed.

The ranking of the objects in order of how they will finish the race;

solid sphere > disk > hollow

The kinetic energy of the objects is calculated as follows;

[tex]K.E = \frac{1}{2} I \omega ^2[/tex]

The ranking of the objects in order of kinetic energy;

solid sphere > disk > hollow

Learn more here:https://brainly.com/question/15076457

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

[tex]\frac{1}{2}mv^{2}=mgh[/tex]

Let's solve it for v:

[tex]v=\sqrt{2gh}[/tex]

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Answer:

95.5 m

Explanation:

The displacement is the position of the ending point relative to the starting point.

In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

         v = (45.83 i ^ -109.56 j ^) m / s

Complete Question:

A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.

(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its

speed is 3.7 m/s, what angle does the cord make with the vertical?

(Check attached image for the diagram.)

Answer:

(a) The ball’s speed, v = 2.06 m/s

(b) The angle the cord makes with the vertical is 50.40⁰

Explanation:

If the ball is revolved in a horizontal plane, it will form a circular trajectory,

the radius of the circle, R = Lsinθ

where;

L is length of the string

The force acting on the ball is given as;

F = mgtanθ

This above is also equal to centripetal force;

[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]

(b) when the speed is 3.7 m/s

[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]

[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]

Therefore, the angle the cord makes with the vertical is 50.40⁰

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ([tex]u^{2}[/tex]*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g

the maximum distance also depends upon square of the initial velocity.

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