Question

A rock thrown vertically upward from the surface of the moon at a velocity of
24 m/sec reaches a height of s=24t-0.8t^2 meters in t seconds.

a) Find the rock’s acceleration and velocity at time t.
b) How long does it take the rock to reach its highest point?
c) How high does the rock go?
d) How long does it take the rock to reach half its maximum height?
e) How long is the rock aloft?

Answer 1

The gravity of the moon is 1.6 m/s * s

Knowing this, you can solve this problem by using different garden-style geomechanics for g:

Vo = 24 m / s

A = -1.6 m/s *sec

The signs of these terms indicate that upward is positive and downward is negative.

The speed at any given time t is given by:

V = Vo + A * t = (24-1.6 * t) m / s

The position at any given time t is given by:

X = Vo * t + A * t ^ 2 = (24 * t-(1/2 * 1.6 * t ^ 2)) m

To find the highest point, request the solution V = 0, because at the highest point, it will stand still for an infinite moment:

24-1.6t = 0

t = 24 / 1.6 = 15 seconds.

To figure out how high the value is, just insert the time into the equation at position X:

X = 24 * t-1.6 * t ^ 2 = 24 * 15-0.5 * 1.6 * 225 = 360 m-180 m = 180 m

If it takes 15 seconds to reach the peak, it also takes 15 seconds to reach the lowest point, so it takes up to 30 seconds. You can also just solve for position X = 0, but you will get two solutions: 0 and 30 seconds. This makes sense: it starts at the bottom (t=0s), then rises, falls, and reaches the bottom again after 30 seconds.

24t-0.8t ^ 2 = 90

0.8t ^ 2 -24t + 90 = 0

8t ^ 2 -240t + 900 = 0

t = 4.3933