Hello,

Solution:

s = ut + 1/2 at^2 is the formula used to find the displacement in 8 th sec

's' - displacement of the object with initial velocity 'u' and acceleration 'a' in the time duration 't'

When a body is thrown vertically upward with velocity u,the distance travelled by it in its 7 th and 8 th seconds will be equal only if it is possible to reach it's maximum height at 7th and freely falls under gravity in 8 th sec.

When a body reaches it's maximum height , it's velocity at that point of time becomes zero

Therefore,at the end of 7 th sec ,u = 0

Then, the body starts with u = 0 at the beginning of 8 th sec and falls freely to the ground under gravity

We have to calculate the distance travelled in 8 th sec by the body In the time travel 7th sec till 8 th sec

Time duration = 8-7 = 1sec

Now, u =0m/s , t = 1s , a= g= 10 m/s^2

We have to substitute these values in the below equation

The displacement of the body in a time interval is given by s= ut+ 1/2 at^2 = 0*1+1/2g*1^2

= 0+1/2 10*1 = 1/2*10 = 5

Hence, the displacement of the body in the 8 th sec is equal to 5 m

So, the value of n is 5

Hope it helps