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Airida [17]
3 years ago
12

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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Answer:

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x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}   (1)

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x_{0}: is the initial position in the horizontal direction = 0

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b) The height of the hill is given by:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final position in the vertical direction = 0

y_{0}: is the initial position in the vertical direction =?

v_{0_{y}}: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)            

g: is the acceleration due to gravity = 9.81 m/s²

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y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m  

I hope it helps you!

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