Question

from the top[ of a tower 45 m high two stones are released. one vertically downwards and othr with a horizontal velocity of 30m/sec . how long will each stone take to strike the ground and how far from tower will each stone strike the ground

Answer 1

Dear Student

Let us consider a body A which is released vertically downwards and another body B projected horizontally with a velocity 30 m/s from the same height and at the same time.
The body B possesses simultaneously: Uniform horizontal velocity 30 m/s and A non-uniform vertical velocity v.
As the body B travels down its vertical velocity (v) increases due to acceleration due to gravity. But the horizontal velocity u remains constant. Hence the body A which is freely falling and the body B projected horizontally from the same height at the same time will strike the ground simultaneously at different points.

Thus time taken by bodies to reach to ground is
$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times 45}{10}}=3sec$

Now the stone dropped vertically will touch the ground in the same spot that is horizontal distance will be zero for body A.

The stone that is thrown horizontally or we can say body B will continue to move with constant horizontal speed of 30 m/s, since there is no component of force and hence acceleration in horizontal direction. So it will strike in
$dis\tan ce=speed\times time=30\times 3=90m$ from the tower.


Regards