Question

A body released from the top of a tower of height H m. After 2s its is stopped and then instantaneously released. What will be its height in 2s

Answer 1

Answer:

G=V-U/T

9.8=0-U/2SEC

U=4.9m/s

USING SECOND EQUATION OF MOTION

S=UT+1/2AT²

A=G

S=4.9+1/2×9.8×2²

S=4.9+1/2×9.8×4

S=4.9+4.9×4

S=9.8×4

S=3.92 MTR

THE HIEGHT =3.92 METER

Explanation:

U=INITIAL VELOCITY

V=FINAL VELOCITY

G=9.8M/S

GIVEN:

V=0

U=?

G=9.8

T=2 SEC