A robot probe drops a camera off the rim of

a 437 m high cliff on Mars, where the free-fall
acceleration is 3.7 m/s
2
.
Find the velocity with which it hits the
ground.
Answer in units of m/s

vf^2=2*a*h

tried that...

vf^2=2*3.7*437
vf^2=3233.8
vf=56.86651036

but my teacher says its the wrong answer....

To find the velocity with which the camera hits the ground, we can use the kinematic equation for free fall:

v^2 = u^2 + 2as

Where:
v = final velocity (what we are trying to find)
u = initial velocity (0 m/s since the camera is dropped from rest)
a = acceleration due to gravity (3.7 m/s^2)
s = displacement or height (437 m)

Since the camera is dropped from rest, the initial velocity (u) is 0. Therefore, the equation simplifies to:

v^2 = 0 + 2 * 3.7 * 437

Now let's solve for v:

v^2 = 2 * 3.7 * 437

v^2 = 3235.6

v = sqrt(3235.6)

v ≈ 56.9 m/s

Therefore, the velocity with which the camera hits the ground is approximately 56.9 m/s.