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3 years ago

6

Josh rolled a bowling ball down a lane in 2.5 seconds. The ball travelled at a constant acceleration of 1.8 m/s/s down the lane

and was traveling at a speed of 7.6 m/s by the time it reached the pins at the end of the lane. How fast was the ball going when it left Josh’s hand?

1 answer:

IRISSAK [1]3 years ago

6 0

<span> Ok, so slope of the lane gives the ball a constant acceleration of 1.8 m/s for 2.5 s
[lane slope = arcsine (1.8/9.8) = 10.6°]
which increases speed of the ball by
1.8(2.5) = 4.5 m/s
The ball reaches the pins with a speed = 7.6 m/s, so it must have left Josh's hand with a speed of
7.6 - 4.5 = 3.1 m/s ANS</span>

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$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

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$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

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