## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 151: 66

#### Answer

The speed of the projectile must be $68.4~m/s$ when it is fired from the cannon.

#### Work Step by Step

We can find an expression for the time of flight: $t = \frac{x}{v_x} = \frac{x}{v_0~cos~\theta}$ We can find the initial speed of the projectile: $y = v_{0y}t+\frac{1}{2}a_yt^2$ $y = v_0~sin~\theta~(\frac{x}{v_0~cos~\theta})+\frac{1}{2}a_y(\frac{x}{v_0~cos~\theta})^2$ $y = x~tan~\theta+\frac{1}{2}a_y(\frac{x}{v_0~cos~\theta})^2$ $y - x~tan~\theta = \frac{a_y~x^2}{2~(v_0~cos~\theta)^2}$ $v_0^2 = \frac{a_y~x^2}{2~(cos~\theta)^2~( y - x~tan~\theta)}$ $v_0 = \sqrt{\frac{a_y~x^2}{2~(cos~\theta)^2~( y - x~tan~\theta)}}$ $v_0 = \sqrt{\frac{(-9.80~m/s^2)(350~m)^2}{2~(cos~40.0^{\circ})^2~(75.0~m - (350~m)~tan~40.0^{\circ})}}$ $v_0 = 68.4~m/s$ The speed of the projectile must be $68.4~m/s$ when it is fired from the cannon.

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