Physics Question?

2015-09-01 4:46 pm

更新1:

A tennis ball is hit straight up at 20.0m/s from the edge of a sheer cliff. Sometime later, the ball passes the original height from which it was hit. (a) How fast is the ball moving at that time? (b) If the cliff is 30.0m high, how long will it take the ball to reach the ground below from the time it was first hit? (c) What total distance did the ball travel? Ignore the effects of wind resistance.

回答 (1)

胡雪8°

2015-09-01 5:19 pm

✔ 最佳答案

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 = (-20)t + (1/2)(10)t²
5t² - 20t = 0
t² - 4t = 0
t(t - 4) = 0

As t ≠ 0, then time taken, t = 4 s

(b)
When the ball reaches the ground :
30 = (-20)t + (1/2)(10)t²
30 = -20t + 5t²
6 = -4t + t²
t² - 4t - 6 = 0

As t > 0, then time taken, t = 2 + √10 = 5.16 s

(c)
When the ball is going up :
v² = vₒ² + 2as
0² = (-20)² + 2(10)s
20s = -400
s = -20 m

Total distance travels = 20 × 2 + 30 = 70 m

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