Question

A particle is projected with a velocity 49 m/s at an angle of 60 degree to the horizontal. calculate the maximum height.

Answer 1

Answer:90  Explanation: H=u²sin²Ф/2g = ( 49²×3)/4×2×10 ≈90

Answer 2

The particle is under projectile motion

Max height = u^2 sin theta / g

Where u is initial velocity

Theta is angle of projection

g is acceleration due to gravity

Max height = 49 * 49 * sin 60 / 10

= 208 m approx