Question

A ball is dropped on the floor from a height of 2.5m. if the ball is in contact with the floor for 0.02s it's average acceleration during contact is

Answer 1

Answer:

a=353.56ms-²

Explanation:

Let the speed be v

then v=√2gh

therefore v=√2*10*2.5

v=√50 ms-¹

now this v will act as initial speed for 0.02s and final velocity is 0 therefore

0=v-at

v=at

a=v/t

a=√50/0.02

a=353.56 ms-²(approx.)

plzz mark it as brainliest