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3 years ago

A ball thrown horizontally at 18.5 m/s from the roof of a building lands 38.9 m from the base of the building.

Physics

1 answer:

Pachacha [2.7K]3 years ago

8 0

Answer: height = 60.5 m

Explanation:

given that the

Initial velocity U = 18.5 m/s

Range = 38.9 m

Let us use second equation of motion under gravity

S = Ut + 1/2gt^2

Where S = range R

Since the motion is horizontal, g = 0

The equation is reduced to:

R = Ut

Make t the subject of formula

t = R/U

Substitute range R and U into the formula

t = 38.9/ 18.5

t = 2.1 s

To calculate the vertical height, we will consider g = 9.81 m/s^2 by using the same second equation of motion

h = Ut + 1/2gt^2

Substitute U, t and g into the formula

h = (18.5 × 2.1) + 1/2 × 9.8 × 2.1^2

h = 38.9 + 21.609

h = 60.51 m

The building is 60.51 metres tall.

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