Question

a stone is left from the top of a building of 40 m so that it reaches the ground vertically in free fall. At the same time another particle is thrown in the horizontal direction from the top of the building which will eventually come down to the ground.If both the particles are released at the same time, then which of them will reach the ground first ??

Answer 1

1. In case of free fall :
a = 9.8 m/s²
u = 0 m/s
h = 40 m
Applying second eq. of motion :
h = u t² + 1/2 a t²
40 = 1/2 X 9.8 X t²
t = 2.85 s
where t is the time in which the stone will reach the earth.

2. In the case of projectile given horizontal projectile :
u = 0 m/s
h = 40 m
Time in which particle will reach the ground is given by the formula :
$T=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 40}{9.8}}=2.85s$

Since, the initial vertical velocity of both the bodies in the two cases is zero and both are accelerated vertically downwards by equal acceleration, hence, both the stone and the particle will reach the ground simultaneously.