## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the time for the ball to travel a horizontal distance of 7.0 meters. $t = \frac{d}{v_x} = \frac{7.0~m}{(20~m/s)~cos(5.0^{\circ})}$ $t = 0.351~s$ We can find the vertical position at this time. $y = y_0+v_{0y}t+\frac{1}{2}at^2$ $y = 2.0~m+(20~m/s)~sin(5.0^{\circ})(0.351~s)-\frac{1}{2}(9.80~m/s^2)(0.351~s)^2$ $y = 2.0~m$ Since the net is 1.0 meter high, the ball clears the net by 1.0 meter.