Question

A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the velocity with which the camera hits the ground.

Answer 1

Answer:

42.05 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 239 m

Acceleration due to gravity (g) = 3.7 m/s²

Final velocity (v) =?

The velocity with which the camera hits the ground can be obtained as follow:

v² = u² + 2gh

v² = 0² + 2 × 3.7 × 239

v² = 0 + 1768.6

v² = 1768.6

Take the square root of both side

v = √(1768.6)

v = 42.05 m/s

Therefore, the velocity with which the camera hits the ground is 42.05 m/s