455 = (1/2)3.7 t^2
t = 15.7 s
v = at = 3.7 * 15.7 = 58 m/s
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
1 answer