Question

A projectile is shot at a velocity of `5 m/s` and an angle of `pi/4`. What is the projectile's peak height?

Answer 1

`0.638m`

Explanation:

Using equations of motion for constant linear acceleration in the vertical direction we get

`v^2=u^2+2ax`

`therefore x=(v^2-u^2)/(2a)`

`=(0^2-(5sin45^@)^2)/(2xx9.8)`

`=0.638m`.