Question

A ball is thrown obliquely upward with the initial speed 12m / s and with the elevation angle 42 degrees. what speed does the ball have after 0.18s?

Answer 1

Horizontal component is

$$v_x=v \cos{42}$$

Vertical component is

$$v_y=v \sin{42}-gt$$

Thus,

$$v=\sqrt{(12\cos{42})^2+(12\sin{42}-(9.8)(0.18))^2}=11\frac{m}{s}$$