Answer
(a) The stone rises to a maximum height of $21.1~m$ above the ground.
(b) The total time that elapses until the stone hits the ground is $4.08~s$
Work Step by Step
(a) At maximum height, we can find the displacement $\Delta y$ above the initial vertical position of $1.50~m$:
$v_f^2 = v_0^2+2a\Delta y$
$\Delta y = \frac{v_f^2-v_0^2}{2a}$
$\Delta y = \frac{0-(19.6~m/s)^2}{(2)(-9.80~m/s^2)}$
$\Delta y = 19.6~m$
The stone rises to a maximum height of $19.6~m + 1.50~m$ which is $21.1~m$ above the ground.
(b) We can find the time to reach maximum height:
$v_f = v_0+at$
$t = \frac{v_f-v_0}{a}$
$t = \frac{0-19.6~m/s}{-9.80~m/s^2}$
$t = 2.0~s$
We can find the time it takes to drop from the maximum height to the ground:
$\Delta y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2\Delta y}{a}}$
$t = \sqrt{\frac{(2)(-21.1~m)}{-9.80~m/s^2}}$
$t = 2.08~s$
The total time that elapses until the stone hits the ground is $2.0~s+2.08~s$ which is $4.08~s$
