## College Physics (4th Edition)

We can convert the horizontal displacement to units of meters: $\Delta x = 45~yd\times \frac{0.9144~m}{1~yd} = 41.148~m$ We can find the time for the ball to travel this horizontal distance: $t = \frac{d}{v_x}$ $t = \frac{d}{v_0~cos~\theta}$ $t = \frac{41.148~m}{(21~m/s)~cos~35^{\circ}}$ $t = 2.392~s$ We can find the ball's vertical displacement at this time: $\Delta y = v_y~t+\frac{1}{2}a_yt^2$ $\Delta y = v_0~sin~\theta~t+\frac{1}{2}a_yt^2$ $\Delta y = (21~m/s)~sin~35^{\circ}~(2.392~s)+\frac{1}{2}(-9.80~m/s^2)(2.392~s)^2$ $\Delta y = 0.776~m$ When the ball has traveled a horizontal distance of 45 yards, the ball is only 0.776 meters above the ground, which is too low to make it over the 10-foot high goal post. Therefore, the ball will not clear the 10-foot high goal post.