Question

A projectile is shot at a velocity of `23 m/s` and an angle of `pi/6`. What is the projectile's peak height?

Answer 1

If `alpha` is the angle of projection with velocity of projection u then vertical component of this velocity is `usinalpha` and final velocity of this component will be zero
so
`0^2= u^2sin^2alpha-2gh`
and
Peak height `h =(u^2xxsin^2alpha)/(2g)=(23^2xxsin^2(pi/6))/(2xx9.8)=6.75m`