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A professional golfer is examining a video of a practice swing. The high‑speed footage shows that his club is in contact with the ball (which was initially at rest on the tee) for only Δ????=0.971Δt=0.971 ms, and the radar gun clocks the speed of the ball as ????b=161vb=161 mph after it comes off the club. The golf ball has a mass of ????b=45.9mb=45.9 g. What is the magnitude of the impulse imparted to the ball by the club?

Respuesta :

Answer:

The impulse imparted to the ball by the club is 3.30 kg m/s.

Explanation:

Given that,

Time = 0.971 ms

Final speed of ball =161 mph

mass of ball = 45.9 g

We need to calculate the impulse

Impulse is equal to the change in momentum

Using formula of impulse

[tex]J=m(v_{2}-v_{1})[/tex]

Where, m = mass of ball

[tex]v_{2}[/tex]= final velocity of ball

Put the value into the formula

[tex]J=0.0459(71.9734-0)[/tex]

[tex]J=3.30\ kg. m/s[/tex]

Hence, The impulse imparted to the ball by the club is 3.30 kg m/s.

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