Question

A projectile is shot at a velocity of `21 m/s` and an angle of `pi/8`. What is the projectile's peak height?

Answer 1

`=62.1m`

Explanation:

If the projectile is shot with velocity u at an angle of projection `alpha` with the horizontal, then it will have initial vertical component of velocity `usinalpha` and this velocity will becomes zero at maximum height H
So Equation of motion can be wrtten as
`0^2=(u*sinalpha)^2-2*g*H`
`H=(u*sinalpha)^2/(2*g)=21^2sin^2(pi/8)/(2*9.8)=62.1m`