A-particle-is-projected-inside-the-tunnel-which-is-4m-high-if-the-initial-speed-is-V-o-show-that-the-maximum-range-inside-the-tunnel-is-given-by-R-4-2-V-o-2-g-8-

Question Number 154547 by peter frank last updated on 19/Sep/21

A particle is projected inside the tunnel

which is 4 m high . if the initial speed

is V_{o} . show that the maximum

range inside the tunnel is given

by R = 4 \sqrt{2} \sqrt{\frac{V_{o}^{2}}{g} - 8}

Answered by peter frank last updated on 20/Sep/21

Commented by peter frank last updated on 20/Sep/21

Range = \frac{{2 v}_{o}^{2} \sin θ \cos θ}{g}

H_{\max} = \frac{v_{o}^{2} \sin^{2} θ}{g} = 4 m

\sin^{2} θ = \frac{8 g}{v_{o}^{2}}

\sin θ = \frac{2 \sqrt{2 g}}{v_{o}}

\cos^{2} θ = \sqrt{1 - \sin^{2} θ}

\cos θ = \frac{1}{v_{o}} \sqrt{v_{o}^{2} - 8 g}

Range = \frac{{2 v}_{o}^{2} \sin θ \cos θ}{g}

Range = \frac{{2 v}_{o}^{2} \frac{2 \sqrt{2 g}}{v_{o}} \frac{1}{v_{o}} \sqrt{v_{o}^{2} - 8 g}}{g}

R = \frac{4 \sqrt{2 g}}{g} \sqrt{v_{o}^{2} - 8 g}

R = 4 \sqrt{2} \frac{\sqrt{g}}{g} \sqrt{v_{o}^{2} - 8 g}

but \frac{\sqrt{g}}{g} = \sqrt{\frac{g}{g^{2}}}

R = 4 \sqrt{2} . \sqrt{\frac{g}{g^{2}} v_{o}^{2} - 8 g}

R = 4 \sqrt{2} . \sqrt{\frac{1}{g} . v_{o}^{2} - 8 g}

R = 4 \sqrt{2} . \sqrt{\frac{v_{o}^{2}}{g} - 8}

Commented by mr W last updated on 20/Sep/21

w e l l s o l v e d!

Commented by peter frank last updated on 21/Sep/21

thank you .

Commented by Tawa11 last updated on 21/Sep/21

nice sir

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