A squash ball hits a wall perpendicularly at 17ms^-1 and rebounds at 10ms^-1 along its
original path . If it takes 0.8s for the ball to rebound, what is the magnitude of the
average acceleration of the ball ?
phy (F.4) easy question
2006-11-03 1:58 am
回答 (1)
2006-11-03 2:21 am
✔ 最佳答案
Let the original direction be positive,average acceleration=(final velocity- intial velocity)/t
a=(v-u)/t
then the answer is (17-(-10))/0.8=33.75ms^-2
收錄日期: 2021-04-23 16:25:02
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