## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $y = v_{0y}t+\frac{1}{2}a_yt^2$ $y = (30~m/s)~sin(60^{\circ})(7.5~s)-\frac{1}{2}(9.80~m/s^2)(7.5~s)^2$ $y = -81~m$ The projectile hits the ground 81 meters lower than the launch point. (b) We can find the maximum height above the ground. $y = \frac{v_y^2-v_{0y}^2}{2g}$ $y = \frac{0-[(30~m/s)~sin(60^{\circ})]^2}{(2)(-9.80~m/s^2)}$ $y = 34~m$ The projectile rises to a maximum height of 34 meters above the launch point.