Question-209098

Question Number 209098 by Spillover last updated on 01/Jul/24

Answered by A5T last updated on 02/Jul/24

4=((v_0 ^2 sin^2 θ)/(2g))⇒sinθ=((√(8g))/v_0 ) R=v_0 cosθ×2t=v_0 (√(1−((8g)/v_0 ^2 )))×2t=2t(√(v_0 ^2 −8g)) R is maximum when t is maximum 4=((gt^2 )/2)⇒t=(√(8/g)) ⇒R=2(√8)×(√((v_0 ^2 /g)−8))=4(√2)((√((v_0 ^2 /g)−8)))

$$\mathrm{4}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\mathrm{2}{g}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{8}{g}}}{{v}_{\mathrm{0}} } \\ $$$${R}={v}_{\mathrm{0}} {cos}\theta×\mathrm{2}{t}={v}_{\mathrm{0}} \sqrt{\mathrm{1}−\frac{\mathrm{8}{g}}{{v}_{\mathrm{0}} ^{\mathrm{2}} }}×\mathrm{2}{t}=\mathrm{2}{t}\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{8}{g}} \\ $$$${R}\:{is}\:{maximum}\:{when}\:{t}\:{is}\:{maximum} \\ $$$$\mathrm{4}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{t}=\sqrt{\frac{\mathrm{8}}{{g}}} \\ $$$$\Rightarrow{R}=\mathrm{2}\sqrt{\mathrm{8}}×\sqrt{\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}−\mathrm{8}}=\mathrm{4}\sqrt{\mathrm{2}}\left(\sqrt{\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}−\mathrm{8}}\right) \\ $$

Commented by Spillover last updated on 03/Jul/24

$$ \\ $$that's right, the question will be wrong after writing 8, they wrote g

Answered by Spillover last updated on 03/Jul/24

H_(max) =(v_y ^2 /(2g)) v_(y ) =vsin θ H_(max) =(((vsin θ)^2 )/(2g)) H_(max) ≤4 metres (((vsin θ)^2 )/(2g)) ≤4 sin^2 θ≤((8g)/v^2 ) sin^2 θ+cos^2 θ=1 cos θ=(√(1−((8g)/v^2 ))) From Range R=((v^2 sin 2θ)/g) R=((v^2 2sin θcos θ)/g) R=((v^2 2(√((((8g)/v^2 )))) .((√(1−((8g)/v^2 ))) ))/g) R=((v^2 2 .((√(((8g)/v^2 )(1−((8g)/v^2 ))) ))/g) R=((v^2 2 .((√(((8g)/v^2 )(((v^2 −8g)/v^2 ))) ))/g) R=((2(√(8g ))(√(v^2 −8g)))/g) ....... A particle is projected with initial speed V at an angle θ to the horizontal. The maximum height H of the projectile must be less than or equal to the height of the tunnel, which is 4.0 meters.

$${H}_{{max}} =\frac{{v}_{{y}} ^{\mathrm{2}} }{\mathrm{2}{g}}\:\:\:\:\:\:\:{v}_{{y}\:} ={v}\mathrm{sin}\:\theta\:\:\:\:\:\:{H}_{{max}} =\frac{\left({v}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}\:\: \\ $$$${H}_{{max}} \leqslant\mathrm{4}\:\:{metres} \\ $$$$\frac{\left({v}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}\:\:\leqslant\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:^{\mathrm{2}} \theta\leqslant\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }}\:\: \\ $$$${From}\:{Range}\: \\ $$$${R}=\frac{{v}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{{g}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{R}=\frac{{v}^{\mathrm{2}} \mathrm{2sin}\:\theta\mathrm{cos}\:\theta}{{g}} \\ $$$${R}=\frac{{v}^{\mathrm{2}} \mathrm{2}\sqrt{\left(\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }\right)}\:.\left(\sqrt{\mathrm{1}−\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }}\:\:\right)}{{g}} \\ $$$${R}=\frac{{v}^{\mathrm{2}} \mathrm{2}\:.\left(\sqrt{\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }\right.}\:\:\right)}{{g}} \\ $$$${R}=\frac{{v}^{\mathrm{2}} \mathrm{2}\:.\left(\sqrt{\frac{\mathrm{8}{g}}{{v}^{\mathrm{2}} }\left(\frac{{v}^{\mathrm{2}} −\mathrm{8}{g}}{{v}^{\mathrm{2}} }\right.}\:\:\right)}{{g}} \\ $$$${R}=\frac{\mathrm{2}\sqrt{\mathrm{8}{g}\:}\sqrt{{v}^{\mathrm{2}} −\mathrm{8}{g}}}{{g}} \\ $$$$……. \\ $$A particle is projected with initial speed V at an angle θ to the horizontal.

The maximum height H of the projectile must be less than or equal to the height of the tunnel, which is 4.0 meters.

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