Answer
(a) The second child reaches a maximum height of 0.82 meters.
(b) The first child's initial speed is 4.8 m/s.
(c) The first child's total time in the air is 0.98 seconds.
Work Step by Step
(a) $y = \frac{v^2-v_0^2}{2a} = \frac{0-(4.0 ~m/s)^2}{(2)(-9.80 ~m/s^2)} = 0.82 ~m$
The second child reaches a maximum height of 0.82 meters.
(b) The first child reaches a maximum height that is 1.5 times the second child's maximum height.
$y = (1.5)(0.82 ~m) = 1.2 ~m$
We can use the first child's maximum height to find the first child's initial speed.
$v_0^2 = v^2 - 2ay = 0 -2ay = -2ay$
$v_0 = \sqrt{-2ay} = \sqrt{(-2)(-9.80 ~m/s^2)(1.2 ~m)} = 4.8 ~m/s$
The first child's initial speed is 4.8 m/s.
(c) We can use the initial speed to find the time it takes to reach the maximum height.
$t = \frac{v-v_0}{a} = \frac{0 -4.8 ~m/s}{-9.80 ~m/s^2} = 0.49 ~s$
The total time in the air is 2t which is 0.98 seconds.
