A robot probe drops a camera off the rim of
a 323 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.
h = 323 m.
V = ?.
To find the velocity with which the camera hits the ground, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity, which we need to find
- u is the initial velocity, which is 0 in this case (since the camera is dropped from rest)
- a is the acceleration due to gravity, which is the free-fall acceleration on Mars, 3.7 m/s^2
- s is the displacement, which is the height of the cliff, 323 m
Let's plug in these values into the equation and solve for v:
v^2 = 0 + 2 * 3.7 * 323
v^2 = 2 * 3.7 * 323
v^2 = 2384.6
Taking the square root on both sides to solve for v:
v = √(2384.6)
v ≈ 48.83 m/s
Therefore, the velocity with which the camera hits the ground is approximately 48.83 m/s.