Question

A projectile is shot at a velocity of `2 m/s` and an angle of `pi/8`. What is the projectile's peak height?

Answer 1

The projectile's peak height is `23.69m`

Explanation:

You have to consider the two component of motion
`s(x)=vt` with `v=vcos22.5°`
`s(y)=vt-1/2at^2` with `v=vsin22.5°`
You have to find the moment when the vertical speed is =0, that is the moment of the peak.
So `v=v-at=0` , `t=g/v=12.82s`
Use this value in the equation of horizontal component
`s(x)=(vcos22.5°)*12.82=23.69m`