Question

a freely falling body travels of distance in 5th second

Answer 1

u = 0. t = 5s
s = u t + 1/2 g t^2
= 0*5 + 1/2* 9.8 * 5^2
= 122.5 m

Answer 2

The distance travelled by the free falling body in 5th second is 122.5 metres.

Given:

That the body is falling freely then the initial velocity of the will be 0 at time t = 5s.

To find:

We have to find the distance travelled by the body which is falling freely.

Solution:

u = 0

t = 5s

We have,

$s=u t+\frac{1}{2} g t^{2}$

Substituting the values of u, t and g in the above expression we get,

Here, g=gravity=9.8

$=0 \times 5+\frac{1}{2} \times 9.8 \times 5^{2}$

$=0+\frac{245}{2}$

$=\frac{245}{2}$

= 122.5 m

Hence the freely falling body with time 5s falls at a distance of 122.5 m.