Question

A particle is projected vertically upwards . It attains a height h after 2 seconds and again after 10 seconds . The speed of the particle at the height h is numerically equal to

Answer 1

Let the desired velocity be u.As the particle returns to the same point in (10−2)(10−2) i.e 8 seconds , its displacement in 8 s is zero.Using s=ut+12s=ut+12at2at2, we get, 0=u(8)−120=u(8)−12g(8)2g(8)2u=4gu=4gHence 4g is the correct answer.

Answer 2

Given,

A particle is projected vertically upwards

It attains a height h after 2 seconds and again after 10 seconds

To Find,

The speed of the particle at the height h =?

Solution,

Let the velocity be u

After 2 seconds, the particle attains a height h then returns to it after 10 seconds

Total time to return to same height = 10 - 2 = 8s

From the second equation of motion, we have

s = ut + 1 / 2gt ²

0 = u* 8 - 1/2g(8)²

8u = 32g

u = 4g

Hence, the speed of the particle at the height h is 4g