Question

Two tall buildings face each other and are at a distance of 180 m from each other. With what velocity must be a ball be thrown horizontally from a window 55m above the ground in one building so that it enters a window 10.9m above the ground in the second building

Answer 1

Vertical distance between the windows is = 540 – 15 = 525 m

Horizontal distance between the windows is = 20 m

Suppose the ball is thrown horizontally with speed ‘u’. Suppose it takes time ‘t’ to reach the other window.

Now,

The ball has no initial velocity along vertical direction. It travels 525 m in time ‘t’.

So, S = ut + ½ at²

=> 525 = 0 + ½ (9.8)(t²)

=> t = 10.35 s

The horizontal component remains unchanged, so, we can use, x = ut

=> 20 = u(10.35)

=> u = 1.93 m/s

Thus, the ball has to be thrown with speed 1.93 m/s horizontally.