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Mariana [72]
3 years ago
5

A cannonball is fired on flat ground

Physics
1 answer:
algol [13]3 years ago
3 0

hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

V₀ = 420m/s and θ₀ = 53.0°

So, when the cannonball is fired it has horizontal and vertical components:

V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s

V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s

When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

Vy = V₀y - g tₐ = 0

tₐ = V₀y/g

tₐ = (335.43m/s)/(9.8m/s²) = 34.23s

Then, the maximum height is reached in the instant tₐ = 34.23s:

h = V₀y tₐ - 1/2g tₐ²

hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²

hmax = 11481.77m - 5741.29m

hmax = 5740.48m

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To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

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sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

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sin ( 90 - 25 ) / sin( 90 - C)  = μ

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From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

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μ = 1.33 / sinC

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