Question

the stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25 m/s calculate when and where the two stones will meet?​

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the two stones will meet.

$\large\underline{ \underline{ \sf \maltese{ \:Given:- }}}$

• Height of tower=100m
• Initial velocity for stone B=25m/s

$\large\underline{ \underline{ \sf \maltese{ \: To\:Find:- }}}$

• When and where the two stones will meet.

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

Let the two stone meet at point 'A' after time 't'.

For Stone A,

u'=0

$\sf{x=0+{\dfrac{1}{2}}gt^2}$

$\sf{x={\dfrac{1}{2}}\times 10\times t^2}$

$\boxed{\orange{\sf{x=5t^2 \dots\dots\dots(1)}}}$

For stone B,

u=25m/s

$\sf{100-x=ut-{\dfrac{1}{2}}gt^2}$

$\boxed{\blue{\sf{100-x=25t-5t^2 \dots\dots\dots(2)}}}$

Adding (1) and (2),

$\sf{x+100-x=5t^2+25t-5t^2}$

$\sf{100=25t}$

$\boxed{\green{\sf{\implies}{t=4s}}}$

From (1),

$\sf{x=5\times 4^2}$

$\boxed{\sf{\red{x=80m}}}$