Question

A stone is dropped from the top of a tower 50m high. At the same time another stone is thrown up from the foot of tower with u=25m/s. When the ball cross?

Answer 1

Answer:

Let stone A be a freely falling body

It's initial velocity (u) = 0 m/s

It's acceleration (a) = 9.8 m/s²

Tower height (h) = 50 m

Calculating the distance travelled by stone A

(From top of a tower)

$\boxed{\rm s = ut + \frac{1}{2}at^2 }$

$\implies \rm s_1 = (0)t + \dfrac{1}{2}at^2$

$\implies \rm s_1 = \dfrac{1}{2}at^2$

For the same stone A, the distance travelled from ground is

$\implies \rm s = 50 - s_1$

$\implies \rm 50 - \dfrac{1}{2} at^2$

Let stone B  thrown up from foot of tower,

It's initial velocity (u) = 25 m/s

It's acceleration (a) = -a (retardation of stone)

Calculating the distance travelled by stone B

(From foot of tower)

$\boxed{\rm s = ut + \dfrac{1}{2}at^2 }$

$\implies \rm s = (25)t + \dfrac{1}{2}(-a)t^2$

$\implies \rm s = 25t - \dfrac{1}{2}at^2$

To find the meeting point of stone A and stone B from the foot of tower

We should equate the distances travelled by stone A and B

Therefore,

$\rm 50 - \dfrac{1}{2} at^2 = \rm 25t - \dfrac{1}{2}at^2$

$\implies \rm 50 - \dfrac{1}{2}at^2 \! \! \! \! \! \! \! \! \! \! \! {\Huge\text{\textbackslash}}} \; = \; \rm 25t - \dfrac{1}{2}at^2 \! \! \! \! \! \! \! \! \! \! \! {\Huge\text{\textbackslash}}}$

♢ Like terms gets cancelled ♢

$\implies \rm 25t = 50$

$\implies \rm t = 2 \;seconds$

From this we can calculate the distance from ground

$\implies \rm 50 - \dfrac{1}{2} at^2$

$\implies \rm 50 - \dfrac{1}{2} (9.8)(2)^2$

$\implies \rm s = 19.6 \;m$

$\blue{\boxed{\boxed{\rm s = 19.6\;m}}}$

Therefore it takes 2 seconds for the stones to cross each other at a distance 19.6 m from foot of the tower.