Question

A tennis ball is hit straight up at 10 m/s from the edge of a sheer cliff. Some time later, the ball passes the original height from which it was hit. (Ignore the effects of air resistance.) (a) How fast is the ball moving at that time? m/s (b) If the cliff is 40 m high, how long will it take the ball to reach the ground level? s (c) What total distance did the ball travel? m Supporting Materials Pcast Pcast Pcast

Answer 1

Take downward quantities to be positive.
Take g = 10 m/s²


(a)
When the ball passes the original height :
s = vot + (1/2)gt²
0 = (-10)t + (1/2)(10)t²
5t² - 10t = 0
t² - 2t = 0
t(t - 2) = 0

As t ≠ 0, then time taken, t = 2 s


(b)
When the ball reaches the ground :
40 = (-10)t + (1/2)(10)t²
40 = -10t + 5t²
8 = -2t + t²
t² - 2t - 8 = 0

As t > 0, then time taken, t = 4 s


(c)
When the ball is going up :
v² = vₒ² + 2as
0² = (-10)² + 2(10)s
20s = -100
s = -5 m

Total distance travels = 5 × 2 + 40 = 50 m