Answer
(a) The initial speed of the baseball was 23 m/s.
(b) The baseball reaches an altitude of 27 m.
(c) The baseball was thrown 0.92 seconds before it passed the window.
(d) The ball reaches the street again 3.7 seconds after passing the window.
Work Step by Step
(a) $v^2 = v_0^2 + 2ay$
$v_0 = \sqrt{v^2 - 2ay}$
$v_0 =\sqrt{(14 ~m/s)^2 - (2)(-9.8 ~m/s^2)(18 ~m)}$
$v_0 = 23 ~m/s$
The initial speed of the baseball was 23 m/s.
(b) $y = \frac{v^2 - v_0^2}{2a}$
$y = \frac{0 - (23 ~m/s)^2}{(2)(-9.8 ~m/s^2)}$
$y = 27 ~m$
The baseball reaches an altitude of 27 m.
(c) $t = \frac{v-v_0}{a} = \frac{14 ~m/s - 23 ~m/s}{-9.8 ~m/s^2} = 0.92 ~s$
The baseball was thrown 0.92 seconds before it passed the window.
(d) We can calculate the time it takes to reach maximum height.
$t = \frac{v-v_0}{a} = \frac{0 - 23 ~m/s}{-9.8 ~m/s^2} = 2.3 ~s$
The total time in the air is 2t which is 4.6 seconds. After passing the window, the time to reach the street again is 4.6 s - 0.92 s = 3.7 seconds.
