Answer
(a) The projectile hits the ground 81 meters lower than the launch point.
(b) The projectile rises to a maximum height of 34 meters above the launch point.
Work Step by Step
(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.
