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Question Number 177548 by peter frank last updated on 07/Oct/22

A projectile is fired with velocity(v_o ) such that it passes through two points both a distance(h) above the horizontal.show that if the gun is adjusted for the maximum range of the separation of two position is d=((v_o (√(v_o ^2 −4gh)))/g)

$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{with}\:\mathrm{velocity}\left(\mathrm{v}_{\mathrm{o}} \right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{two} \\ $$$$\mathrm{points}\:\mathrm{both}\:\mathrm{a}\:\mathrm{distance}\left(\mathrm{h}\right)\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{horizontal}.\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{the}\:\mathrm{gun}\:\mathrm{is}\:\mathrm{adjusted} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{two}\:\mathrm{position}\:\mathrm{is} \\ $$$$\boldsymbol{\mathrm{d}}=\frac{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} \sqrt{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} ^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{gh}}}}{\boldsymbol{\mathrm{g}}} \\ $$

Answered by blackmamba last updated on 07/Oct/22

y=y_o +x tan θ −((gx^2 )/(2v_o ^2 cos^2 θ)) since θ=45° ,y=h , y_o =0 then h=0+x tan 45°−((gx^2 )/(2v_o ^2 cos^2 45°)) b= x−((gx^2 )/v_o ^2 ) or x^2 −(v_o ^2 /g).x +(v_o ^2 /g)h =0 x_1 +x_2 = (v_o ^2 /g) ⇒(x_1 −x_2 )^2 = (x_1 +x_2 )^2 −4x_1 x_2 ⇒d^2 = ((v_o ^2 /g))^2 −4(((v_o ^2 h)/g)) ⇒d=(√(((v_o ^2 /g))^2 (v_o ^2 −4gh))) ⇒d =((v_o (√(v_o ^2 −4gh)))/g)

$${y}={y}_{{o}} +{x}\:\mathrm{tan}\:\theta\:−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}_{{o}} ^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \theta}\: \\ $$$${since}\:\theta=\mathrm{45}°\:,{y}={h}\:,\:{y}_{{o}} =\mathrm{0} \\ $$$${then}\:{h}=\mathrm{0}+{x}\:\mathrm{tan}\:\mathrm{45}°−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{v}_{{o}} ^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{45}°} \\ $$$${b}=\:{x}−\frac{{gx}^{\mathrm{2}} }{{v}_{{o}} ^{\mathrm{2}} }\:{or}\:{x}^{\mathrm{2}} −\frac{{v}_{{o}} ^{\mathrm{2}} }{{g}}.{x}\:+\frac{{v}_{{o}} ^{\mathrm{2}} }{{g}}{h}\:=\mathrm{0} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\:\frac{{v}_{{o}} ^{\mathrm{2}} }{{g}} \\ $$$$\Rightarrow\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\:\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\Rightarrow{d}^{\mathrm{2}} =\:\left(\frac{{v}_{{o}} ^{\mathrm{2}} }{{g}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{v}_{{o}} ^{\mathrm{2}} \:{h}}{{g}}\right) \\ $$$$\left.\Rightarrow{d}=\sqrt{\left(\frac{{v}_{{o}} ^{\mathrm{2}} }{{g}}\right)^{\mathrm{2}} \left({v}_{{o}} ^{\mathrm{2}} −\mathrm{4}{gh}\right.}\right) \\ $$$$\Rightarrow{d}\:=\frac{{v}_{{o}} \:\sqrt{{v}_{{o}} ^{\mathrm{2}} −\mathrm{4}{gh}}}{{g}} \\ $$

Commented by Tawa11 last updated on 07/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by Spillover last updated on 07/Oct/22

x=v_o cos θt y=v_o sin θt−(1/2)gt^2 R=((v_o ^2 sin 2θ)/g) θ=(π/4) x=((v_o t)/( (√2))) y=((v_o t)/( (√2)))−(1/2)gt^2 x^2 −((v_o ^2 x)/g)+((v_o ^2 h)/g)=0 x_1 =(v_o ^2 /(2g))−(v_o /(2g))(√(v_o ^2 −4gh)) x_2 =(v_o ^2 /(2g))−(v_o /(2g))(√(v_o ^2 −4gh)) d=x_2 −x_1 d=(v_o /g)(√(v_o ^2 −4gh)) ★★★

$$\mathrm{x}=\mathrm{v}_{\mathrm{o}} \mathrm{cos}\:\theta\mathrm{t}\:\:\mathrm{y}=\mathrm{v}_{\mathrm{o}} \mathrm{sin}\:\theta\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\mathrm{R}=\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{g}}\:\:\:\:\:\:\:\:\:\:\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{x}=\frac{\mathrm{v}_{\mathrm{o}} \mathrm{t}}{\:\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\mathrm{y}=\frac{\mathrm{v}_{\mathrm{o}} \mathrm{t}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{x}}{\mathrm{g}}+\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{h}}{\mathrm{g}}=\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} }{\mathrm{2g}}−\frac{\mathrm{v}_{\mathrm{o}} }{\mathrm{2g}}\sqrt{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} −\mathrm{4gh}}\: \\ $$$$\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} }{\mathrm{2g}}−\frac{\mathrm{v}_{\mathrm{o}} }{\mathrm{2g}}\sqrt{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} −\mathrm{4gh}}\: \\ $$$$\mathrm{d}=\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} \\ $$$$\mathrm{d}=\frac{\mathrm{v}_{\mathrm{o}} }{\mathrm{g}}\sqrt{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} −\mathrm{4gh}}\:\:\:\:\bigstar\bigstar\bigstar \\ $$$$ \\ $$

Commented by peter frank last updated on 07/Oct/22

more clarification please step 3? and how θ=(π/4)? diagram please?

$$\mathrm{more}\:\mathrm{clarification}\:\mathrm{please}\:\mathrm{step}\:\mathrm{3}? \\ $$$$\mathrm{and}\:\mathrm{how}\:\theta=\frac{\pi}{\mathrm{4}}?\:\mathrm{diagram}\:\mathrm{please}? \\ $$$$ \\ $$

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