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A 1.30 kg ball is attached to a ceiling by a

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1 answer:

amid [387]2 years ago

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Put A, It will help you.

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2 years ago

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Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

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t = Time taken = 0.04 seconds

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a = Acceleration

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$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

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The height from which the student fell is 5.7141 m

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