A robot probe drops a camera off the rim of a 387 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2 .
Find the velocity with which it hits the ground.
Answer in units of m/s
V^2 = Vo^2 + 2g*h
Vo = 0
g = 3.7 m/s^2
h = 387 m.
Solve for V
2863.8
To find the velocity with which the camera hits the ground, we can use the formula for the final velocity in free fall:
v = √(2 * g * h)
Where:
v: final velocity of the camera
g: acceleration due to gravity on Mars (3.7 m/s^2)
h: height from which the camera is dropped (387 m)
Plugging in the given values into the formula, we have:
v = √(2 * 3.7 * 387)
Simplifying further:
v = √(2 * 3.7 * 387)
v = √(2 * 3.7 * 387)
v = √(2882.8)
v ≈ 53.7 m/s
Therefore, the velocity with which the camera hits the ground is approximately 53.7 m/s.