Question

Question #ebb84

Answer 1

At max height the projectile is travelling horizontally.

Assuming no resistance, and for launch veclocity `vec v_o` at angle `theta` to horizontal, it's horizontal velocity throughout is `abs (vec v_o) cos theta \ vece_x`.

So we can say that:

`abs (vec v_o) = 5 abs (vec v_o) cos theta`

`implies theta = cos^(-1) \ 1/5`