## College Physics (4th Edition)

(a) We can find the time it takes to reach the ground: $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(20.0~m)}{9.80~m/s^2}}$ $t = 2.02~s$ The ball will fall to the ground in a time of $2.02~s$ (b) Since the initial vertical velocity is zero like part (a), the ball will fall to the ground in a time of $2.02~s$ (c) We can find the time it takes to reach the ground: $y = v_{0y}~t+\frac{1}{2}at^2$ $y = v_0~sin~\theta~t+\frac{1}{2}at^2$ $20.0 = (20.0)~sin~18^{\circ}~t+4.9~t^2$ $4.9~t^2 +6.18~t-20.0=0$ We can use the quadratic formula to find $t$: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-6.18 \pm\sqrt{(6.18)^2 - 4(4.9)(-20.0)}}{(2)(4.9)}$ $t = -2.75~s, 1.49~s$ Since the time must be positive, it takes 1.49 seconds for the ball to reach the ground.