A ball dropped from a tower reaches the ground in 4 s the height of the tower is (take g=10 ms 2)

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As the stone is dropped, initial velocity u = 0.g = 10 m/s^2time t = 4 sec.Distance = ut + 1/2 at^2.

S = 0 + 1/2 * 10 * 8 = 40 m is the height of the tower.

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t=4 seca=10m/sec^2 (due to gravity)u=0m/secs=?so,S=u*t+1/2*a*(t)^2S=0*4+1/2*10*(4)^2S=0+5*16

therefore distance=80m

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A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.

Initial velocity u = 0

Time t = 4 s

g = 10 m/s2

Let 'H' be the height of the tower.

Using the second equation of motion,

H = ut + (1/2) gt2

Or, H = 0 + (1/2)(10)(4)2

Or, H = 80 m

Concept: Newton’s Second Law of Motion

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