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 stone will meet.

Explanation

Here height of the tower is 100 m .Now ,suppose the two stones meet is a point p which is at a point P which is at a height x above the ground as shown in the figure so that the distance of point P from the top of the tower is 100-x .
(i) For the stone falling from top of tower :

Height ,h=(100-x) m
Intitial velocity ,u=0
Time t=?
And Acceleration due `g=9.8 m//s^(2)` (Stone falls towns )
to gravity
Now , `h=ut+1/2gt^(2)`
So, `100-x=0xxt+1/2xx9.8xxt^(2)`
or, `100-x=4.9 t^(2)`
(ii) For stone projected vertically upwards :
Height , h=x m
Initial velocity u=25 m/s
Time t=?
And Acceleration due ,`g=-9.8m//s^(2)` (stone goes up )
Now `s=ut+1/2gt^(2)`
`x=25xxt+1/2xx(-9.8)xxt^(2)`
or, `x=25 t-49 t^(2)`
On adding equations (1) and (2) we get
`100-x+x=4.9t^(2)+25 t - 4.9 t^(2)`
100=25 t
`t=100/25`
t= 4s
thus ,the two stones will meet after a time of 4 second
Now from equation (1) we have
`100-x=4.9 t^(2)`
Putting t=4 in this equations we get
`100-x=4.9xx(4)^(2)`
`100-x=4.9xx 16`
100 -x =78.4
100-78.4=x
21.6 m =x
x=21.6 m
Thus , the two stones will meet at a height of 21.6 meters from the gorund .