Question

a boy is dropped from a 100m high cliff and at the same time another body is thrown up from the ground at 25m/s. where will the two meet?

Answer 1

Hence distance where they meet is 20m

Answer 2

Answer:

The distance x=21.6 m from the ground, both boys will meet with each other.

Explanation:

The total height is 100 m

Let x be the height from the ground at which both boys meet

The distance traveled by the first boy which is thrown from the cliff is

100-x

The initial velocity of the first boy is zero

So by the laws of motion

$s=ut-\dfrac {1}{2}gt^2\\100-x=\dfrac{1}{2}9.8 t^2\\100-x=\dfrac{1}{2}9.8 t^2$                                 (1)

The initial speed of the second boy is u=25 m/s

The distance traveled by the second boy which is thrown from the ground is

$x=25 t-\dfrac{1}{2}9.8 t^2$                     (2)

Putting value of equation 1 in equation 2 we get,

$100-25 t-\dfrac{1}{2}9.8 t^2=\dfrac{1}{2}9.8 t^2\\100-25t=0\\25t =100\\t=4\ s$

Forthe value of x

$x=25 t-\dfrac{1}{2}9.8 t^2\\x=25\times 4-\dfrac{1}{2}9.8 4^2\\x=21.6\ m$