Question

An object is dropped from a height of 100m.its velocity at the moment it touches the ground ?​

Answer 1

Answer:

The answer is 20√5 m/s

Explanation:

PE= mgh

m*10*100

PE=1000m

At moment it touches ground,

PE=KE

by, work energy theorum

KE= 1/2 mv2

1000m= 1/2mv2

2000=V2

v=20√5

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Answer 2

Here we have to find the final velocity of the ball

u=0m/s (dropped from rest with no speed)

v=?

s= -100m (-ve because of sign convention)

a=9.8m/s²

ATQ

v²-u²=2as

=>v²=2as+u²

v²=2×9.8(-100)+0

v²=19.6×-100

v²= -1960

v=±√1960

v= -44.27 (-ve sign shows opposite direction)

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