# Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 13

The speed of the cannon ball when it reaches the ground is 81.2 m/s

#### Work Step by Step

We can use conservation of energy to find the speed $v_f$ of the cannon ball when it reaches the ground. Therefore; $KE_f+PE_f = KE_0+PE_0$ $\frac{1}{2}mv_f^2+0=\frac{1}{2}mv_0^2+mgh$ $v_f^2=v_0^2+2gh$ $v_f= \sqrt{v_0^2+2gh}$ $v_f= \sqrt{(80~m/s)^2+(2)(9.80~m/s^2)(10~m)}$ $v_f = 81.2~m/s$ The speed of the cannon ball when it reaches the ground is 81.2 m/s.

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