Answer
(a) The speed of the stone at the height of 13.0 m is 17.9 m/s.
(b) The stone is at a height of 13.0 m when t = 0.620 s and when t = 4.28 s.
(c) There are two solutions because the stone is at a height of 13.0 m two times, once on the way up (t = 0.620 s) and once on the way back down (t = 4.28 s).
Work Step by Step
(a) $v^2 = v_0^2 + 2ay$
$v = \sqrt{(24.0 ~m/s)^2 - (2)(9.80 ~m/s^2)(13.0 ~m)}$
$v = 17.9 ~m/s$
The speed of the stone at the height of 13.0 m is 17.9 m/s.
(b) $y = v_0t + \frac{1}{2}at^2$
$13.0 = 24.0\cdot t - 4.90\cdot t^2$
$4.90 \cdot t^2 - 24.0 \cdot t + 13.0 = 0$
We can use the quadratic formula to solve for $t$.
$t = \frac{-b ~+/- ~\sqrt{b^2 - 4ac}}{2a} = \frac{24.0 ~+/- \sqrt{(-24.0)^2 - (4)(4.90)(13.0)}}{(2)(4.90)}$
$t = 0.620 ~s, ~4.28 ~s$
The stone is at a height of 13.0 m when t = 0.620 s and when t = 4.28 s.
(c) There are two solutions because the stone is at a height of 13.0 m two times, once on the way up (t = 0.620 s) and once on the way back down (t = 4.28 s).
