Question

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to gravity at the surface of the moon in 1.62 m/s2. What is the maximum height of the rock above the surface?

Answer 1

Answer: 277.777 m

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was projected upward from the surface, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$    (1)

$V=V_{o}-g.t$    (2)

Where:

$y$  is the rock's final position

$y_{o}=0$  is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$  is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$    (3)

On the other hand, the maximum height  is accomplished when $V=0$:

$V=V_{o}-g.t=0$    (4)

$V_{o}-g.t=0$    

$V_{o}=g.t$    (5)

Finding $t$:

$t=\frac{V_{o}}{g}$    (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$    (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$    (8)  Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$   (9)

Finally:

$y_{max}=277.777m$