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Gnom [1K]
3 years ago
9

Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of th

e ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? assume that the ground was perfectly flat. express your answer using two significant figures.
Physics
1 answer:
Umnica [9.8K]3 years ago
8 0

horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

x = vcos\theta * t

y = v sin\theta * t - \frac{1}{2} gt^2

solving above two equations we have

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now here we will plug in all data

0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}

0.9 = 150.65 - \frac{184150.2}{v^2}

\frac{184150.2}{v^2} = 149.75

v = 35.1 m/s

<em>so the ball was hit with speed 35.1 m/s from the ground</em>

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Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

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Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

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\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a (2)

Where:

T - Tension force in the cord, measured in newtons.

m_{A}, m_{B} - Masses of blocks A and B, measured in kilograms.

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By subtracting (2) by (1), we get an expression for the acceleration of each mass:

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a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)

a = 4.203\,\frac{m}{s^{2}}

The acceleration of the masses is 4.203 meters per square second.

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T = m_{B}\cdot (a+g)

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t = \sqrt{\frac{2\cdot \Delta y}{a} }

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v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

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