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Rina8888 [55]
3 years ago
5

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

Physics
1 answer:
lozanna [386]3 years ago
7 0
You can find

1) time to hit the ground
2) initial velocity
3) speed when it hits the ground

Equations

Vx = Vxo

x = Vx * t

Vy = Vyo + gt

Vyo = 0

Vy = gt

y = yo - Vyo - gt^2 / 2

=> yo - y = gt^2 / 2

1) time to hit the ground

=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

=> t = √1.631 s^2 = 1.28 s

2) initial velocity

Vxo = x / t = 6.5m / 1.28s = 5.08 m/s

3) speed when it  hits the ground

Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s

V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2

=> V = √ (183.56 m^2 / s^2) = 13.55 m/s
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F = m a

In order to calculate the acceleration, given the displacement d,  

d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

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This value can be used to determine the time for the pellet through the barrel:

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16∠45° Ω

Explanation:

Applying,

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