A robot probe drops a camera off the rim of a 105.0 m high cliff on Mars, where the free-fall acceleration is -3.70 m/s2. Find the velocity in m/s with which the camera hits the ground.

sqrt(2g'H), where g' is the acceleration of gravity on Mars.

sqrt(7.4*105) = 27.9 m/s

No

42 and 11

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

v^2 = u^2 + 2as

Where:
v = final velocity (which is what we need to find)
u = initial velocity (in this case, the initial velocity is zero, as the camera is dropped)
a = acceleration due to gravity (which is -3.70 m/s^2 because it is in the opposite direction on Mars)
s = displacement (the vertical distance traveled, which is 105.0 m in this case)

Plugging in the values, the equation becomes:

v^2 = 0^2 + 2(-3.70)(105.0)

Simplifying further:

v^2 = 2(-3.70)(105.0)

v^2 = -777.00

Since the velocity cannot be negative in this context (as it represents speed), we can take the square root of both sides:

v = √(-777.00)

Now, we have a complex number as the result. However, in the context of physical motion, we are only interested in the positive real solution. So, the velocity in m/s with which the camera hits the ground is approximately 27.87 m/s.