Question

A robot probe drops a camera off a 239 m high cliff on Mars. The free-fall acceleration on Mars is -3.7 m/s². Mars has very little atmosphere, so the camera will fall freely. Find the velocity when the camera hits the ground.

a) 68.3 m/s
b) 34.1 m/s
c) 51.2 m/s
d) 17.0 m/s

Answer 1

Using the formula for free fall, the camera dropped from a cliff on Mars will hit the ground with a velocity of approximately 68.3 m/s, which corresponds to option (a).

Step-by-step explanation:

The velocity when the camera hits the ground on Mars is approximately 68.3 m/s. Using the formula vf = sqrt(2gh), where vf is the final velocity, g is the acceleration due to gravity on Mars (-3.7 m/s²), and h is the height of the cliff (239 m), we can determine the velocity at impact. Substituting the values, we obtain:

vf = sqrt(2 * (-3.7 m/s²) * 239 m) = sqrt(-1774.6 m²/s²) = 68.26 m/s

Approximating this value, the camera's velocity when it hits the ground would be 68.3 m/s, corresponding to option (a).

The velocity when the camera hits the ground on Mars is approximately 68.3 m/s. Given that the robot probe drops a camera off a 239 m high cliff on Mars, and the free-fall acceleration on Mars is -3.7 m/s², we can calculate the velocity when the camera hits the ground using the formula: vf = sqrt(2gh). Where vf is the final velocity, g is the acceleration due to gravity on Mars (-3.7 m/s²), and h is the height of the cliff (239 m). Plugging in these values, we get: vf = sqrt(2 * (-3.7 m/s²) * 239 m) = 68.26 m/s. Therefore, the velocity when the camera hits the ground is approximately 68.3 m/s.