Question

A cannonball shot at 51 m/s at an angle of 38°. What will be the maximum height that the cannonball reaches? Units?

Answer 1

The maximum height reached by the cannonball can be found as follows:

$$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(51\ \dfrac{m}{s})^2\cdot sin^238^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=50.3\ m.$$