A tennis ball, starting from rest, rolls down a hill. At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin walled spherical shell (I=2/3 mr^2 )and determine the range x it travels if it starts at a height of 2.50 m

To determine the range of the tennis ball, we need to find the horizontal distance it travels before landing.

Let's break down the problem into multiple steps:

Step 1: Find the velocity of the ball as it leaves the ground.
Given that the ball started from rest and rolled down the hill, the initial vertical velocity can be determined using the conservation of energy. The potential energy it had at the top of the hill is converted into kinetic energy at the bottom of the hill. Therefore:

(mgh) initial = (1/2)mv^2

Where:
m = mass of the ball
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the hill

Let's assume the mass of the ball is 0.05 kg and the height of the hill is 2.50 m. Plugging in these values, we can solve for the velocity:

(0.05 kg)(9.8 m/s^2)(2.50 m) = (1/2)(0.05 kg)v^2

v^2 = (0.05 kg)(9.8 m/s^2)(2.50 m) / (1/2)(0.05 kg)
v^2 = 24.5 m^2/s^2
v = √(24.5 m^2/s^2)
v ≈ 4.95 m/s

So, the velocity of the ball as it leaves the ground is approximately 4.95 m/s.

Step 2: Decompose the velocity into horizontal and vertical components.
The horizontal component of the velocity (Vx) remains constant throughout the projectile motion, while the vertical component (Vy) changes due to gravity. We can determine the horizontal and vertical components of the velocity using the angle given.

Vx = v * cosθ
Vy = v * sinθ

Where:
θ = angle of projection (35° in this case)

Plugging in the known values:

Vx = 4.95 m/s * cos 35°
Vx ≈ 4.06 m/s

Vy = 4.95 m/s * sin 35°
Vy ≈ 2.82 m/s

So, the horizontal component of velocity (Vx) is approximately 4.06 m/s, and the vertical component (Vy) is approximately 2.82 m/s.

Step 3: Determine the time of flight.
The time of flight (T) is the total time the ball is in the air. We can calculate it using the vertical component of velocity.

Using the kinematic equation:

Vy = gt - 1/2 * g * t^2

Where:
g = acceleration due to gravity (approximately 9.8 m/s^2)

Rearranging the equation to solve for t:

1/2 * g * t^2 - gt + Vy = 0

Plugging in the known values:

1/2 * (9.8 m/s^2) * t^2 - (9.8 m/s^2) * t + 2.82 m/s = 0

Solving this quadratic equation using the quadratic formula:

t = [-(-9.8 m/s^2) ± √((-9.8 m/s^2)^2 - 4 * 1/2 * (9.8 m/s^2) * 2.82 m/s)] / (2 * 1/2 * (9.8 m/s^2))

t = [9.8 m/s^2 ± √(96.04 m^2/s^2 - 27.996 m^2/s^2)] / 9.8 m/s^2

t = [9.8 m/s^2 ± √68.044 m^2/s^2] / 9.8 m/s^2

t = (9.8 m/s^2 ± 8.258 m/s) / 9.8 m/s^2

t1 ≈ 1.00 s
t2 ≈ 2.29 s

So, the time of flight is approximately 1.00 s or 2.29 s. However, we choose the larger value as the ball leaves at an angle of 35° and therefore will spend more time in the air.

Step 4: Calculate the range.
The total horizontal distance covered by the ball is calculated by multiplying the horizontal component of velocity (Vx) with the time of flight (T).

Range (x) = Vx * T

Plugging in the known values:

Range (x) ≈ 4.06 m/s * 2.29 s

Range (x) ≈ 9.32 m

Therefore, the tennis ball will travel a range of approximately 9.32 meters.

To determine the range x that the tennis ball travels, we can use the principles of projectile motion.

First, let's break down the problem into two parts: the horizontal motion and the vertical motion of the ball.

1. Vertical Motion:
Since the ball starts from rest and becomes airborne at an angle of 35° with respect to the ground, we can use the equation of motion in the y-direction:
y = yo + vot + (1/2)at^2

Since the initial vertical velocity is 0 (starting from rest), this equation simplifies to:
y = (1/2)at^2

Given that the initial height is 2.50 m, we have:
2.50 = (1/2) * (-9.8) * t^2 (taking acceleration due to gravity as -9.8 m/s^2)

Solving for time, we get:
t^2 = -5.10 / -9.8
t^2 = 0.5204

Taking the positive square root, we find:
t = 0.722 seconds

2. Horizontal Motion:
The horizontal range x is given by:
x = vox * t

To find the horizontal velocity vox, we can use the conservation of angular momentum. For a thin-walled spherical shell, the angular momentum is given by Iω, where I is the moment of inertia and ω is the angular velocity.

Given that the moment of inertia (I) for a thin-walled spherical shell is (2/3) * m * r^2, where m is the mass of the ball and r is the radius, and that angular momentum is conserved, we can equate the initial and final angular momentum.

The initial angular momentum is zero since the ball starts from rest, and the final angular momentum is given by:
I * ω = m * v * r

Simplifying, we have:
(2/3) * m * r^2 * ω = m * v * r

Simplifying further, we find:
ω = (3/2) * (v / r)

Since ω = Δθ / Δt (change in angle / change in time), and the initial angle is 0° and the final angle is 35°, we can write:
35° / t = (3/2) * (v / r)

Rearranging, we get:
v = r * (35° / t) * (2/3)

Substituting the known values:
v = 0.35 * (2.50 / 0.722) * (2/3)
v ≈ 3.609 m/s

Now, substituting the values of v and t into the equation for horizontal range, we get:
x = 3.609 * 0.722
x ≈ 2.61 m

Therefore, the tennis ball travels a horizontal distance of approximately 2.61 meters.

You need how far it rolled down the hill to get initial speed

v = w r where w is omega, the angular velocity

kinetic energy = m g h
= (1/2)I w^2 + (1/2) mv^2

= (1/2)(2/3)mr^2v^2/r^2 + (1/2)mv^2

=(mv^2/2)[2/3 + 1] = (5/3) (mv^2/2)

so
g h = (5/3)(v^2/2)

solve for v, initial launch speed
Vi = v sin 35
u = v cos 35

then it is conventional parabolic flight problem.