Question

An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds. The acceleration due to gravity on that planet is

Answer 1

it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m

Answer 2

Answer:

Acceleration due to gravity on that planet, a = 9 m/s²

Explanation:

Initial velocity, u =  0 m/s

Acceleration due to gravity , a = ?

Displacement, s = 72 m

Time , t = 4 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    72 = 0 x 4 + 0.5 x a x 4²

   a = 9 m/s²

Acceleration due to gravity on that planet, a = 9 m/s²