Answer
(a) It takes 2.02 seconds for the ball to reach the ground.
(b) It takes 2.02 seconds for the ball to reach the ground.
(c) It takes 1.49 seconds for the ball to reach the ground.
Work Step by Step
(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.
