Question

A cannon tilted up at a ${30}^0$angle fires a cannon ball at $80m/s$from a top a $10-m$-high fortress wall. What is the ball’s impact speed on the ground below?

Short answer

Ball's impact speed on the ground$v_i=69m/s$

Step-by-step solution

Given information

speed of cannon ball $v=80m/s$

inclination angle of cannon $\theta ={30}^0$

height of the wall$h=10m$

Explanation

Resolving the speed of cannon ball into horizontal and vertical component.

$$\begin{gathered} v_x=v\cos \left(\theta \right) \\ {v}_y=v\sin \left(\theta \right) \end{gathered}$$

Therefore, the ball impact speed on ground is given by horizontal component of speed.

$$\begin{gathered} v_i=v\cos \left(\theta \right) \\ \Rightarrow v_i=80\times \cos \left({30}^0\right)=69.3\approx 69m/s \end{gathered}$$