The acceleration due to the moon's gravity at its surface is less than what it is here on the surface of the earth. The computed value of 1.6 m/s2 is consistent with what the Apollo Program astronauts observed during their explorations there. Imagine an astronaut holding a piece of moon rock over the rim of a crater. At time zero the rock is 500 m above the crater floor and is tossed upward at 5.0 m/s. What will be the velocity of this rock 15 s after it was tossed upward?
g=1.6 m/s²
Upward motion
v=v₀-gt
v=0
t= v₀/g=5/1.6 =3.125 s
Downward motion
t₁=15-3.125 =11.875 m/s
v=gt₁=1.6•11.875=19 m/s
To find the velocity of the rock 15 seconds after it was tossed upward, we can use the kinematic equation:
v = u + at
Where:
v is the final velocity (what we're trying to calculate),
u is the initial velocity (5.0 m/s),
a is the acceleration due to the Moon's gravity (-1.6 m/s^2 due to being tossed upward), and
t is the time in seconds (15 s).
First, let's determine the initial velocity of the rock when it reaches the highest point of its trajectory. At that point, the velocity will be zero. So, we can write:
v = u + at
0 = 5 + (-1.6)t
-5 = -1.6t
Solving for t:
t = -5/(-1.6)
t ≈ 3.125 s
So, it takes approximately 3.125 seconds for the rock to reach its highest point.
Now, let's find the final velocity of the rock after 15 seconds. Since 15 seconds is greater than the time it takes to reach the highest point, we'll need to consider the downward acceleration due to the Moon's gravity. In this case, the total time will be the time to reach the highest point plus the additional time of 15 seconds. So, the total time will be 3.125 s + 15 s = 18.125 s.
Using the kinematic equation:
v = u + at
v = 0 + (-1.6)(18.125)
v ≈ -29 m/s
Therefore, the velocity of the rock 15 seconds after it was tossed upward will be approximately -29 m/s. The negative sign indicates that the rock is moving downward.