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3 years ago

5

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo

nent of the ball? Round the answer to the nearest tenth of a m/s.

1 answer:

marishachu [46]3 years ago

6 0

To calculate the horizontal velocity component, use the cosine identity. where cos A = a / h
where A is the angle the golfer hits the ball
a is the horizontal component velocity
h is the speed the golfer hits the ball

so a = h cos A
a = 31 cos 35
a = 25.4 m/s

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

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The figure of the problem is missing: find in attachment.

(a) 1.64 s

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$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

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3 years ago