Answer
(a) h = 2.8 m
(b) h = 1.5 m
(c) h = 1.5 m
Work Step by Step
We can use conservation of energy to solve this question. The energy at height $h$ is equal to the energy as the Lunar Module lands. To find the maximum possible height $h$, let's assume that the Lunar Module is moving at a speed of 3.0 m/s when it lands.
(a) $PE_1 = KE_2$
$mgh = \frac{1}{2}mv^2$
$h = \frac{v^2}{2g} = \frac{(3.0 ~m/s)^2}{(2)(1.62 ~m/s^2)} = 2.8 ~m$
(b) $PE_1 + KE_1 = KE_2$
$mgh + \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2$
$mgh = \frac{1}{2}m(v_2^2 - v_1^2)$
$h = \frac{(v_2^2 - v_1^2)}{2g} = \frac{(3.0 ~m/s)^2- (2.0 ~m/s)^2}{(2)(1.62 ~m/s^2)} = 1.5 ~m$
(c) $PE_1 + KE_1 = KE_2$
$mgh + \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2$
$mgh = \frac{1}{2}m(v_2^2 - v_1^2)$
$h = \frac{(v_2^2 - v_1^2)}{2g} = \frac{(3.0 ~m/s)^2- (2.0 ~m/s)^2}{(2)(1.62 ~m/s^2)} = 1.5 ~m$
