Question

A cannonball is shot from the top of a hill 7 meters above the field below with an initial speed of 100m/s and a launch angle of 50 degrees. What is the maximum height the cannon ball reached above the field?

Answer 1

Let's first find the time that the cannonball takes to reach its maximum height:

$$v_y=v_{0}sin\theta-gt,$$ $$0=v_{0}sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

The maximum height that the cannonbal can reach above the field can be found as follows:

$$y_{max}=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the previous formula we can find the maximum height that the cannonbal can reach above the field:

$$y_{max}=y_0+v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=y_0+\dfrac{v_0^2sin^2\theta}{2g}=7\ m+\dfrac{(100\ \dfrac{m}{s})^2\cdot sin^250^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=306.4\ m.$$