Time Needed To Reach The Ground

What was the problem? 

To solve the problem, we need to remember what we have learned in our early days of school. The gravity of the earth accelerates the falling object at the rate of 9.8 meter per second square. We know, the kinematic equation,

s = ut + (1/2) a t^2   .......(1)

where,

s = distance covered,
u = initial velocity,
a = acceleration,
t = time taken to travel distance s.

In this case, initial velocity must be 0 as it is dropped from height 54m. Again height here is the distance covered by the object. And the acceleration in this case is nothing but the acceleration due to gravity which is 9.8 m/s^2.

So putting s = 54 m, u = 0 m/s, a = 9.8 m/s^2 in equation (1) above,

54 = 0 x t + (1/2) x 9.8 x t^2

54 = 4.9 t^2

t^2 = 11.021

t = 3.3 seconds. 

So theoretically both balls should take 3.3 seconds to reach the ground. But resistance due to air will make the difference in time taken by balls to hit the ground.