Question

A stone is dropped from the top of a tower 50m high.at the same time another stone is thrown upwardfrom the foot of the tower with a velocity of 25m/s .when and where the two stone cross each other .

Answer 1

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Answer 2

Answer:

Both the stone will meet at a distance of 30.4 m from the foot of the tower after 2 sec.

Explanation:

Given that,

A stone is dropped from the top of a tower 50 m high.at the same time another stone is thrown upward from the foot of the tower with a velocity of 25 m/s .

Total height of tower = 50 m

let both stone meet at a height h from the foot of the tower

Using equation of motion

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

Where, u = initial velocity

v = final velocity

s = height

g = acceleration due to gravity

Put the value into the equation

$50-h=0+\dfrac{1}{2}\times9.8\times t^2$

$50-h=4.9t^2$....(I)

For second stone

Using equation of motion again

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

$h =25t-\dfrac{1}{2}\times9.8\times t^2$

$h = 25t-4.9t^2$....(II)

On adding equation (I) and (II)

$50=25t$

$t =2 sec$

Now, put the value of t in equation (I)

$50-h=4.9\times4$

$h = 50-4.9\times4$

$h = 30.4\ m$

Hence, Both the stone will meet at a distance of 30.4 m from the foot of the tower after 2 sec.