Question

A body is freely falling from a height of 98 m. During its fall, after 2 sec acceleration due to gravity is disappeared. The velocity of striking of the body​

Answer 1

First we need to find out the distance traveled in 2 sec than we will find the velocity when gravity is disappeared.

As the gravity disappears the velocity will become constant.

H=ut+gt²/2

H=0+20

H=20m

Means body has travelled 20m in 2 sec

now the velocity after 2 sec→

v=u+at

v=20m/s.

hence 20m/s is the velocity when acceleration disappears

hence it will strike with velocity 20m/s

Answer 2

Explanation:

Use v²-u² = 2as

Here a = acceleration due to gravity = g(freely falling)

s = height dropped( distance covered)

u = 0( no force applied on body it's just falling freely)

So, v² = 2gh

→ v² = 2×10×98 → v = 14√10 ms^-1