Question

A ball when released from a height of 44 .1 m ,the velocity with which the ball strikes the           ground when g = 9.8 m/s2  ​

Answer 1

Answer:

velocity = 29.4m/s

Explanation:

initial speed of the stone (u) =0 m/s

let the final speed be v

height of the tower (H) =44.1 m

acceleration due to gravity (g) = 9.8 m/s ^²

now using the second equation of motion

H = UT + ½ gt²

44.1 =0 +½×9.8 × t²

t = 44.1

4.9

t= 3 second

v=u+at

v = 0 + 9.8

v=29.4 m/s