Question

Question #45954

Answer 1

`theta ~~ 42^@`

Explanation:

Note that it takes `30/2 = 15` seconds for the vertical component of the velocity to become `0`.

Denoting the initial vertical component of the velocity by `v` and using `g = 9.8 ms^(-2)`, we have:

`v = 9.8 * 15 = 147`

So the launch angle `theta` satisfies:

`sin(theta) = 147/220`

So `theta = sin^(-1)(147/220) ~~ 42^@`