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kifflom [539]
2 years ago
14

A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

Physics
1 answer:
sveticcg [70]2 years ago
8 0

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h      conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m       max height

Check:

t = 28.2 / 9.8 = 2.88 sec    time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

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Answer:

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Explanation:

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Given:

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\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1

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F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2

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\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times  R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

\mu_s=\frac{1}{3}\tan \theta

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