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Mike is serving the volleyball for the second time in a volleyball game. If the ball leaves his hand with twice the velocity it had on the first serve, its horizontal range R would be:

Respuesta :

oladayoademosu

Answer:

His new horizontal range is 4 times his initial range.

Explanation:

Since Mike serves the ball with velocity, u, his horizontal range is

R = u²sin2Ф/g where Ф is the angle between u and the horizontal.

Now, if on his second serve, the ball leaves his hand with twice the velocity of his initial serve, the new velocity is v = 2u.

So, the new range R' = v²sin2Ф/g

R' = (2u)²sin2Ф/g

R' = 4u²sin2Ф/g

Since R = u²sin2Ф/g,

R' = 4u²sin2Ф/g

R' = 4R

So, his new horizontal range is 4 times his initial range.

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