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Masja [62]
3 years ago
11

A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio

n of the kick starts running towards the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

U = √Rg/sin2θ

Explanation:

Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.

Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.

According to the range formula,

R = U²sin2θ/g

Cross multiplying we have;

Rg = U²sin2θ

Dividing both sides by sin2θ, we have;

U² = Rg/sin2θ

Taking the square root of both sides we have;

√U² = √Rg/sin2θ

U = √Rg/sin2θ

Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ

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Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%.  (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

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Answer:

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Explanation:

It is given that,

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