Question

A particle is thrown with velocity u making an
angle with the vertical. It just crosses the top
of two poles each of height h after 1 s and 3s
respectively. The maximum height of projectile is

Answer 1

Answer:

Explanation:

tH=v0yg and

H=v20y2g=(gtH)22g=12gt2H

Here t H=(1+3)2=2s

Hence H=12(9.8)(2)square =19.6m