Q. A BODY RELEASED FROM THE TOP OF A TOWER FALLS THROUGH HALF THE HEIGHT OF THE TOWER IS 2s. IN WHAT TIME SHALL THE BODY FALL THROUGH THE HEIGHT OF THE TOWER.

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Solution

s = ut + ½at²
u = 0 , a = 9.8 m/s²
s = 0 + ½ 9.8 2²

or

s = 19.6 m
h = 2s = 39.2 m

Again,
h = 0 + ½at²
39.2 = ½ 9.8 t²
t² = 8
t = 2âˆš2 seconds

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A body dropped from the top of a tower falls through 40m during the last 2s of its fall. The height of the tower is ( $g = 10 m / s^{2}$ )

A body dropped from top of a tower fall through $40 m$ during the last two seconds of its fall. The height of the tower is
( $g = 10 {m/s}^{2}$ )

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(g = 10 m s^{- 2})

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