Physics, asked by serena394, 5 months ago

its urgent please give answer correctly ​

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Answered by Anonymous
47

\large\sf { t = \sqrt{ \frac{2h}{g}}}

\large\sf { \ and \ mw^{2} R = \mu mg}

\large\sf { \ so \ w = \sqrt { \frac{ \mu g}{R}}}

\large\sf { v = wr = \sqrt{R \mu g}}

so range \large\sf { = \sqrt{R \mu g} \times \sqrt{ \frac{2h}{g}}}

\large\sf { = \sqrt{ 2 \mu Rh}}

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