. A 0.660 kg ball is dropped from rest from the top of a building and falls for
5.65 seconds. How tall was the building? [Neglect friction. Assume on or near
the Earth's surface.]

Answer :

Answer:

The height of the building is 156.42 m

Step-by-step explanation:

Given;

mass of the ball, m = 0.66 kg

time of fall, t = 5.65 s

initial velocity of the ball, u = 0

Apply the following kinematic equation to determine the height of the building;

h = ut + ¹/₂gt²

h = ¹/₂gt²

h = ¹/₂ x 9.8 x 5.65²

h = 156.42 m

Therefore, the height of the building is 156.42 m

The building was 156.42 metres tall

From the question we are told that

the mass of the ball, m = 0.66 kg

the time of fall, t = 5.65 s

the initial velocity of the ball, u = 0

If we use one of the equations of motions in

h = ut + ¹/₂gt², we then have

h = (0 * t) +  (0.5 * 9.8 * 5.65²)  

h = 4.9 * 31.923  

h = 156.42 m

The height of the building is 156.42 m

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