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2 years ago

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A paintball is fired horizontally from a tower 45 m above the ground. If the paintball gun fires at 90 m/s… How long does it tak

e for the paintball to hit the ground? How far away from the tower will the paintball land?

1 answer:

ankoles [38]2 years ago

3 0

Answer:

The time taken for the paint ball to hit the ground is $t = 2.143 \ s$

The distance of the landing point from the tower is $d = 192.86 \ m$

Explanation:

From the question we are told that

The height of the tower is $h = 45 \ m$

The speed of the paintball in the horizontal direction is $v_x = 90 \ m/s$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} at^2$

Here u is the initial velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally

a is equivalent to $g = 9.8\ m/s^2$

t is the time taken for the paintball to hit the ground

So

$45 = 0* t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.143 \ s$

Generally the distance of its landing position from the tower is

$d = v * t$

=> $d = 90 * 2.143$

=> $d = 192.86 \ m$

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Answer:

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Explanation:

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A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8

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Answer:

$v = 186.90\,\frac{m}{s}$

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

$\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h$

The final velocity of the system formed by the ballistic pendulum and the bullet is:

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$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}$

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Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

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Answer:

a) There are 100 centimeters in 1 meter.

b) $\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

Explanation:

a) We have the conversion

1 m = 100 cm

So there are 100 centimeters in 1 meter.

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$\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

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Answer:

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