On the moon, the acceleration due to gravity is -1.6 meters per second per second. A stone is dropped from a cliff on the moon and hits hits the surface of the moon 20 seconds later. How far did it fall? What was its velocity at impact?
s = 1/2 at^2
v = at
plug in t=20, a = -1.6
45 is wrong
To determine the distance the stone fell on the moon, we can use the equation of motion:
d = (1/2) * g * t^2
Where:
- d is the distance the stone fell
- g is the acceleration due to gravity (-1.6 m/s^2 on the moon)
- t is the time it took for the stone to fall (20 seconds)
Plugging in the values:
d = (1/2) * (-1.6 m/s^2) * (20 s)^2
Calculating this equation gives us:
d = (1/2) * (-1.6 m/s^2) * (400 s^2)
d = -0.8 m/s^2 * 400 s^2
d = -320 m
Therefore, the stone fell approximately 320 meters downwards on the moon.
Now, let's determine the stone's velocity at impact. We can use the equation:
v = g * t
Where:
- v is the velocity
- g is the acceleration due to gravity (-1.6 m/s^2 on the moon)
- t is the time it took for the stone to fall (20 seconds)
Plugging in the values:
v = (-1.6 m/s^2) * (20 s)
v = -32 m/s
Therefore, the stone had a velocity of -32 m/s at the impact on the moon. The negative sign indicates that the velocity was in the opposite direction of the acceleration due to gravity.