Question

a particle is projected vertically upwards and it covers a distance of 10m in its5th second. The maximum height attained by the particle is (g=10m/s2)

Answer 1

That's it first apply formula for distance travelled in n the second

Answer 2

Given question is solved using the formula of equation of motion of instantaneous displacement.

                   $S_{ t } = u +\frac{1}{2} a(2t-1)$

Given data are the distance travelled in the 5th second is 10 m. therefore                                          

                   $S_{ 5 } = 10 m.$

Putting the value into the equation, we get,

                   $10 = u + \frac{1}{2}a (2t -1)$

The acceleration is the acceleration due to gravity as the motion is acting downwards will be treated as negative.  

                   $10 = u + \frac{1}{2} (-10)\times (2\times5 -1)$

                   u = 10 + 45 = 55m/s  

The maximum height attained will be at final velocity being zero.  

By,

                   $v^2 -u^2 = 2as$

                   $0 -(55)^2 = 2\times10\times S$

                    s = 151.25m