Lunar Elevation

Let y be the elevation of the projectile, that is to say its position. The gravitational field is the acceleration of the projectile. Velocity is the first time-derivative of position, and acceleration is the second time-derivative of position. This means the second derivative of y equals the gravity.

$\frac{\mathrm d^{2}}{\mathrm d t^{2}} y = a = -1.62$

The position vector y points up and is represented here with positive numbers. The acceleration vector a points down, and thus needs to be represented with a negative number. For the same reason, the velocity is positive when the projectile is moving upward, and is negative when the projectile is moving downward.

To find the velocity you integrate the acceleration

$v = \int a\; \mathrm{d}t = -1.62 \int dt = -1.62 \, t + v_0$

At t=0 the projectile is just leaving the ground and has its initial velocity of 20 m/s.

$v(0) = v_0 = 20$

$v(t) = -1.62 \, t + 20$

To find position you integrate the velocity

$y = \int v\; \mathrm{d}t = -1.62 \int t\; \mathrm{d}t + 20 \int dt = -1.62 \cdot \frac{1}{2} t^{2} + 20 \, t + y_0$

At t=0 the projectile is at its initial position, which is an elevation of 0.

$y(0) = y_0 = 0$

$y(t) = -0.81 \, t^{2} + 20 \, t$

Plugging 20 seconds in for t you get

$y(20) = 76$

20 seconds after its launched the projectile is 76 meters above the surface of the Moon.