Question

A particle is projected in vertically upward from the ground it attains maximum height H in time T what will be its height after time 4/3T​

Answer 1

So the body is projected vertically upward. Then,

Initial Velocity = u [θ = 90°]

Acceleration = - g

Given that the body travelled its maximum height H for a time T, i.e.,

H = uT - (gT² / 2) → (1)

But, we know that the velocity of the particle projected at the highest point is zero. Then, by the first kinematic equation,

u - gT = 0

u = gT

Then (1) becomes,

H = gT² - (gT² / 2)

H = gT² / 2

After a time 4T / 3, let the height of the particle be H'. Then,

H' = u (4T / 3) - (g (4T / 3)² / 2)

H' = (4gT² / 3) - (g (16T² / 9) / 2) [u = gT]

H' = (8 (gT² / 2) / 3) - (16 (gT² / 2) / 9)

H' = (8 H / 3) - (16 H / 9)

H' = 8 H / 9