Question

A body is projected vertically upwards with initial velocity of 25m/s. during third second of its motion what would be the distance travelled

Answer 1

Let us take v as 0 to find out the time at which the body turns and comes towards the ground .
So v will be 0
Hence u/a = t
( as v is equal to zero )
So 25m/s/10m/s^2 = t
Hence the body would stop at 2.5 seconds and would move towards the ground .
So to calculate the distance we would first calculate the distance to the top and then the distance after 0.5 second from the top as 2.5 seconds have already passed in the motion .
So first distance till the top = s = v^2 - u^2/ 2a
S = 25m/s^2/2.10
S= 31.25 till the top
After the top the distance equal to =
S = ut + 1/2at^2
S = 1/2 . 10 . 0.5 s
S = 2.5 m
So total distance equal to 31.75 m