Question

I watched an astronaut drop a feather on the moon. It took 1.3 seconds. How fast was it going right before it hit the lunar surface if freefall accelleration on the moon is 1.62 m/s^2?​

Answer 1

2.106 m/sec

Explanation:

An astronaut drops a feather on the moon and it took 1.3 seconds to reach the surface.

Now, it is given that, the free-fall acceleration on the moon is 1.62 m per sec².

There is a relation for moving body under acceleration due to gravity that,

v = u+ gt .......(1), where, v and u are the final and initial velocities of the falling body under acceleration due to gravity g and t is the time of travel.

Therefore, in our case, u =0 as there is free fall, g =1.62 m/sec² and t= 1.3 secs.

Hence, the velocity of the feather just before hitting the lunar surface(v) will be given by v = 0+ 1.62 ×1.3 =2.106 m/sec.

{Using the formula (1)} (Answer)