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# Another problem from kinematics

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.800 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

let the total height be h

So, for first case

S = h/2

a = g=10 m/s^2

u=0 m/s

t = 0.8 sec

using the relation $S=ut + \frac{1}{2}at^{2}$

h/2 = 0.5 x 10×0.8 ^2

h=6.4 m

In second case (considering the full motion)

S=h=6.4m

t=?

a=g=10m/s^2

u=o m/s

using the relation $S=ut + \frac{1}{2}at^{2}$
6.4 = 0.5 x 10 x t^2

or

t = 2 x 6.4/10 = 1.28 s

## 1 Comment

1. zeen says:

here we have to use Newton’s equations of motion.

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