Question

A body falls freely from rest. It covers as much
distance in the last second of its motion as
covered in the first three seconds. The body has
fallen for a time of
(a) 3s
(b) 5 s
(c) 7s
(d) 9 s​

Answer 1

Given : Body is freely falling (u = 0 ; a = - g)

It covers same distance in the last second of its motion as covered in the first three seconds.

we have for uniformly accelerated bodies,

$s = ut + \frac{1}{2}*at^2$

The distance covered in the first three seconds ,

 $s_1 = 0 + \frac{1}{2} * -g*3^2$      

⇒ $s_1 =- \frac{9}{2} *g$

Let the total time of free fall be ' T '  

so, distance covered in last second,

$s_2 = - \frac{1}{2} *g*T^2 - [ - \frac{1}{2} *g*(T-1)^2]$

⇒   $s_2 = - \frac{1}{2} *g*[(T^2-(T-1)^2]$

⇒ $s_2 = -\frac{1}{2} *g*[2T-1]$

Given, $s_1 = s_2$

⇒ $- \frac{9}{2} *g= -\frac{1}{2} *g*[2T-1]$

⇒ 9 = [2T-1]

⇒ 2T = 10

∴ T = 5 sec

CORRECT OPTION IS : (B) 5 s