Question

A stone is allowed to fall from a tower of height

200 m and at the same time another stone is

projected vertically upwards from the ground at

a velocity of 20 m/s. Calculate when and where

the stones will meet.​

Answer 1

The stone will be meet at 77.5 m in 10 sec.

Explanation:

Given that,

Height = 200 m

Velocity = 40 m/s

We need to calculate the height for first stone  

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$(200-h)=0+\dfrac{1}{2}\times9.8t^2$

$200-h=4.9t^2$

...(I)

For second stone,

$a = -g$

Using equation of motion again

$s=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

....(II)

From equation (I) and (II)

$40t=200$

$t=5 sec$

We need to calculate the value of h

Using equation (I)

$4.9t^2+h=200$

Put the value of t

$4.9\times25+h=200$

$h=200-4.9\times250$

$h=77.5\ m$

Hence, The stone will be meet at 77.5 m in 10 sec.

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Topic : height

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