Question

A body is released from the top of a tower of height 50m find the velocity with which the body hit the ground G is equal to 9.8 metre per second square

Answer 1

Assuming free fall and applying equation of kinematics

v^2 = u^2 + 2gs

Here u^2 = initial speed = 0

v^2 = 2 ×9.8 ×50

v^2 = 980

v = 14 × 2.23

= 31.22 m/s

Answer 2

Answer:

the above answer is correct

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