Question

A projectile is shot at a velocity of `23 m/s` and an angle of `pi/4`. What is the projectile's peak height?

Answer 1

To answer this the formulas `u_y = usin(theta)`, `v_y = u_y + a_yt` and `y = u_yt + (a_yt^2)/2` can be used.

Explanation:

First we want vertical component of velocity
`u_y = usin(theta)`
`u_y = 23sin(pi/4)`
`u_y = 23/sqrt(2)`

Second we find the time taken to reach a vertical velocity of `0` (this is its peak height).
`v_y = u_y + a_yt`
`0 = 23/sqrt(2) -9.8/t`
`t =9.8xxsqrt(2)/23`
(Assuming the projectile is on Earth `a_y = -9.8` )

Finally we can find the peak height
`y = u_yt + (a_yt^2)/2`
`y = 23/sqrt(2) xx 9.8 xx sqrt(2)/23 - 9.8 xx (9.8 xx sqrt(2)/23)^2/2`

`y = 9.8 - 9.8^3/23^2`
`y = 8.02` `m`