Question

2. Sally rolls a ball up to another person on a smooth ramp 19.6 m above her. The ball reaches

the other person's hands when it is travelling 4.9 m/s uphill. If the ramp angle slows the ball
down by 3.7 m/s each second it travels up the ramp, find the initial velocity of the ball.

Answer 1

Take Sally's position to be the origin, and up-the-ramp to be the positive direction. The ball travels a distance x in time t of

x = u t + 1/2 (- 3.7 m/s²) t²

where u is the ball's initial velocity.

Its velocity v at time t is

v = u + (- 3.7 m/s²) t

Let T be the time it takes for the ball to reach the second person 19.6 m up the ramp. At this time, the ball attains a velocity of 4.9 m/s, so that

4.9 m/s = u + (- 3.7 m/s²) T

T = (u - 4.9 m/s) / (3.7 m/s²)

Substitute this into the distance equation, with x = 19.6 m, and solve for u :

19.6 m = u (u - 4.9 m/s) / (3.7 m/s²) + 1/2 (- 3.7 m/s²) ((u - 4.9 m/s) / (3.7 m/s²))²

u ≈ 13 m/s