The girl throws the 0.5-kg ball toward the wall with an initial velocity va = 10 m/s. determine (a) the velocity at which it strikes the wall at b, (b) the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5, and (c) the distance s from the wall to where it strikes the ground at c.

a) 8.81 m/s

b) 4.62 m/s

c) 3.967 meter

Explanation:

I found the complete exercise, there it says the distance between the wall and the position of the ball at t=0; is 3meters. The ball is thrown at an angle of 30°. The ball, at t=0, is at a height of 1.5meters

Let's consider the direction from the ball to the wall as positive.

Step 1: Data given

Mass of the ball =0.5 kg

Initial velocity vA = 10 m/s

Step 2: Calculate the time the ball needs to hit the wall

sx = s0 +vxt

⇒ with sx = the horizontal component of the ball, after the impact

⇒ with s0 = the horizontal component of the ball at time t=0

⇒ with vx = the horizontal component of the velocity ( vx = vA*cos 30°)

⇒ with t = time

sx = s0 + vA*cos 30° *(t)

3 = 0 + 10*cos30° (t)

t = 0.3464 s

Step 3: Calculate the velocity of the ball before touching the wall

Since the ball is thrown in the air, let's consider the upwards direction as positive.

The vertical component of the velocity vy can be defined as followed:

vy = v0y + at

⇒ with vy = the vertical component of velocity

⇒ with v0y = the vertical component of initial velocity (v0y = vA*sin30°)

⇒ with a = the acceleration of the ball (in this case a = -g) (negative because the ball falls)

vy = Va*(sin30°)+at

vy = 10*sin30° -9.81*0.3464

vy = 1.60 m/s

Step 4: Calculate the vertical position of the ball, where the ball touches the wall

The vertical component sy can be written as followed:

sy = s0y +v0y*t + 1/2 at²

⇒ with sy = the vertical component of the ball, where he touches the wall

 ⇒ with s0y = the vertical component of the ball at time t=0

⇒ with v0y = the vertical component of its initial velocity (v0y = vA sin30°)

sy = s0y +vAsin30° *t +1/2 at²

sy = 1.5 + (10sin30° *0.3464) -1/2*9.81 *(0.3464)²

sy =sB = 2.64m

Step 5: Calculate the magnitude of the velocity at the point when and where the ball touches the wall

The magnitude of the velocity at point B can be written as followed:

vB = √(vx² +vy²)

vB = √((vAcos30°)² + vy²)

 ⇒ with vAcos30° = 10*cos 30° = 8.66

⇒ with vy = 1.6

vB = √(8.66²+1.6²)

vB = 8.81 m/s

b) Calculate the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5 Let's call the velocity of the ball when it rebounds vB2.

This velocity has a horizontal component (vB2x) and a vertical component (vB2y)

Step 6: Define the coefficient of restitution e

e = (vB2 - vA2)/(vA1 - vB1)

⇒ vB2x = the horizontal component of vB2 = TO BE DETERMINED

⇒ vA2 = 0

⇒ vA1 = 0

⇒ vB1 = vx = 10cos30°

e = (vB2x - 0)/(0-8.66)

0.5 = (vB2x - 0)/(0-8.66)

vB2x = 4.33 m/s ( in the reverse direction)

vB2y = vBy1 = 1.6 m/s

vB2 = √((vB2x)² + (vB2y)²)

vB2 = √(4.33² + 1.6²)

vB2 = 4.62 m/s

The ball will rebound under an angle of

vB2*cosα = 4.33

cosα = 4.33/4.62

α = 20.4 °

The velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5 ; is 4.62 m/s

c) Calculate the distance s from the wall to where it strikes the ground at C.

sy = s0y + v0y*t +1/2at²

⇒ with sy = sB (but reverse direction, so different sign)

⇒ with v0y = (vB2)*sinα

-sB = 0 + (vB2)*sinα *t1 -1/2*9.81*(t1)²

-2.64 = 0 + 4.62 * sin(20.4°) -1/2*9.81*(t1)²

0 = 4.905(t1)² - 1.610*t1 -2.64

t1 = 0.916 s

Step 7: Calculate the horizontal distance after the impact

sx = s0x +vx*t1

sx = s0x + (vB2)*cosα *t1

sx = 0 + 4.62* cos(20.4)*0.916

sx = 3.967 meter

power out to power in less than 100%