A ball is thrown straight up with enough speed so that it is in the air for seve

Question

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is upwards. a) What is its velocity 1.1 s before it reaches its highest point? b) What is the change in its velocity, v, during this 1.1-s interval? c) What is its velocity 1.1 s after it reaches its highest point? d) What is the change in velocity, v, during this 1.1-s interval? e) What is the change in velocity, v, during the 2.2-s interval from 1.1 s before the highest point to 1.1 s after the highest point? (Caution: We are talking about velocity, not speed.). f) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?

Explanation / Answer

a)

The rate of acceleration at Earth's surface is 9.8 m/s^2. That means that every second the ball's velocity changes by 9.8 meters per second. If it's travelling up, its upward velocity is slowed by 9.8 meters per second. If it's travelling down, it's downward velocity is sped up by 9.8 meters per second. If it changes by 9.8 meters every second, then it's easy to figure out how fast it was moving 1.1 second before it's velocity was 0. Hence, v=9.8m/s^2*1s-9.8m/s^2*0.1s=8.82m/s

b) from 8.82 m/s up to 0 m/s = -8.82 m/s change

c) 1.1 after reaching the top, it is moving down a v = -9.8m/s^2*1.1s=- 10.78 m/s; the minus sign on the velocity indicates that it is, indeed, moving down.

d) from 0 m/s up to 10.78 m/s = 10.78 m/s change

e)v(t=1.1safter) - v(t=1.1sbefore) = 10.78 m/s - 8.82m/s = 1.96m/s

f) g=9.8m/s^2

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