Question

If an object is freely falling from a hugh tower of 100 m and another ball is thrown vertically upwards with a velocity of 25m/s where and when the two objects meet

Answer 1

$<b><i>There are two objects : One is freely falling object and other is a ball thrown vertically upwards$


Using the formula

$S=h + ut+\frac{1}{2}at^{2}$

==\\\[1]///==
Freely Falling object :

It is made to fall from the height h= 100 mts
u=0;
a=g=9.8m/s

$S= 100 + \frac{1}{2}(9.8)t^{2}$

$S = -4.9t^2+100$ ---[1]

==\\\[2]///==
Ball thrown vertically upwards

It is thrown with velocity u= 25 m/s
and from height h= 0 m from ground
a=- g= - 9.8m/s

$S=0 + ut+\frac{1}{2}at^{2}$

$S= 25t + \frac{1}{2}(-9.8)t^{2}$

$S = -4.9t^2+25t$ ---[2]


Equate [1] & [2]
$-4.9t^2+100$ = $-4.9t^2+25t$

25t = 100
t = 4 sec

So these objects meet after 4 seconds

Substitute in [1] to find the position:
$S=-4.9t^2+100=21.6$

For 1 sec ball travels 21.6 mts
But they meet after 4 seconds

The balls meet 4 x 21.6 = 78.4 meters below the top of the tower.