A robot probe drops a camera off the rim of a 455 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
455 = (1/2)3.7 t^2
t = 15.7 s
v = at = 3.7 * 15.7 = 58 m/s
To find the velocity with which the camera hits the ground, we can use the equation for free-fall motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (in this case, the free-fall acceleration on Mars), and s is the distance fallen.
In this case, the camera is dropped from rest, so the initial velocity, u, is 0 m/s. The distance fallen, s, is given as 455 m, and the free-fall acceleration, a, is 3.7 m/s^2.
Plugging in the values into the equation, we get:
v^2 = 0^2 + 2 * 3.7 * 455
Simplifying, we have :
v^2 = 0 + 3367
v^2 = 3367
Now, we can solve for v by taking the square root of both sides:
v = √(3367)
Calculating this, we find that the velocity with which the camera hits the ground is approximately 58.01 m/s. Therefore, the answer is 58.01 m/s.