Question

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

Answer 1

Answer:

tH=v0yg and

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

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

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