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Question Number 16455 by Tinkutara last updated on 22/Jun/17

The speed of a projectile when it is at its greatest height is (√(2/5)) times its speed at half the maximum height. What is its angle of projection?

$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{its}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{is}\:\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{times}\:\mathrm{its} \\ $$$$\mathrm{speed}\:\mathrm{at}\:\mathrm{half}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{its}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}? \\ $$

Answered by sandy_suhendra last updated on 23/Jun/17

A=point of the maximum height B=point of half maximum height V_A = V_0 cos α h_A = ((V_0 ^( 2) sin^2 α)/(2g)) V_A = (√(2/5)) V_B ⇒ V_B = (5/2)(√(2/5)) V_A =(5/2)(√(2/5)) V_0 cos α h_B = (1/2)h_A = ((V_0 ^( 2) sin^2 α)/(4g)) KE_A +PE_A =KE_B +PE_B (1/2)m.V_A ^( 2) + m.g.h_A = (1/2)m.V_B ^( 2) + m.g.h_B (1/2)m(V_0 cos α)^2 + m.g.((V_0 ^( 2) sin^2 α)/(2g)) = (1/2)m((5/2)(√(2/5)) V_0 cos α)^2 + m.g.((V_0 ^( 2) sin^2 α)/(4g)) (1/2) cos^2 α + (1/2) sin^2 α = (5/4) cos^2 α + (1/4) sin^2 α (1/4) sin^2 α = (3/4) cos^2 α tan^2 α = 3 ⇒ tan α = (√3) ⇒ α = 60°

$$\mathrm{A}=\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{B}=\mathrm{point}\:\mathrm{of}\:\mathrm{half}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{V}_{\mathrm{A}} \:=\:\mathrm{V}_{\mathrm{0}} \:\mathrm{cos}\:\alpha \\ $$$$\mathrm{h}_{\mathrm{A}} \:=\:\frac{\mathrm{V}_{\mathrm{0}} ^{\:\:\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{2g}} \\ $$$$\mathrm{V}_{\mathrm{A}} \:=\:\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{V}_{\mathrm{B}} \:\Rightarrow\:\mathrm{V}_{\mathrm{B}} \:=\:\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{V}_{\mathrm{A}} =\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{V}_{\mathrm{0}} \:\mathrm{cos}\:\alpha\:\:\:\:\: \\ $$$$\mathrm{h}_{\mathrm{B}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{h}_{\mathrm{A}} \:=\:\frac{\mathrm{V}_{\mathrm{0}} ^{\:\:\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{4g}} \\ $$$$\mathrm{KE}_{\mathrm{A}} +\mathrm{PE}_{\mathrm{A}} =\mathrm{KE}_{\mathrm{B}} +\mathrm{PE}_{\mathrm{B}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}.\mathrm{V}_{\mathrm{A}} ^{\:\:\mathrm{2}} \:+\:\mathrm{m}.\mathrm{g}.\mathrm{h}_{\mathrm{A}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}.\mathrm{V}_{\mathrm{B}} ^{\:\:\mathrm{2}} \:+\:\mathrm{m}.\mathrm{g}.\mathrm{h}_{\mathrm{B}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left(\mathrm{V}_{\mathrm{0}} \:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \:+\:\mathrm{m}.\mathrm{g}.\frac{\mathrm{V}_{\mathrm{0}} ^{\:\:\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{2g}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left(\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{V}_{\mathrm{0}} \:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \:+\:\mathrm{m}.\mathrm{g}.\frac{\mathrm{V}_{\mathrm{0}} ^{\:\:\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{4g}}\:\:\:\:\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}^{\mathrm{2}} \alpha\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \alpha\:=\:\frac{\mathrm{5}}{\mathrm{4}}\:\mathrm{cos}^{\mathrm{2}} \alpha\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}^{\mathrm{2}} \alpha \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}^{\mathrm{2}} \alpha\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{cos}^{\mathrm{2}} \alpha \\ $$$$\mathrm{tan}^{\mathrm{2}} \alpha\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{tan}\:\alpha\:=\:\sqrt{\mathrm{3}}\:\Rightarrow\:\alpha\:=\:\mathrm{60}° \\ $$

Commented by Tinkutara last updated on 23/Jun/17

But answer is 60°.

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{60}°. \\ $$

Commented by sandy_suhendra last updated on 23/Jun/17

sorry, I′ve a mistake at the second line from the bottom but I′ve corrected

$$\mathrm{sorry},\:\mathrm{I}'\mathrm{ve}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{at}\:\mathrm{the}\:\mathrm{second}\:\mathrm{line}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bottom} \\ $$$$\mathrm{but}\:\mathrm{I}'\mathrm{ve}\:\mathrm{corrected} \\ $$

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by ajfour last updated on 23/Jun/17

At half the max. height: (1/2)mv^2 =(1/2)mu^2 − ((mgH)/2) v^2 =u^2 −((u^2 sin^2 θ)/2) ...(i) Also as given ucos θ =(√(2/5)) v or v^2 = (5/2)u^2 cos^2 θ (ii) equating (i) with (ii) u^2 −((u^2 sin^2 θ)/2) = (5/2)u^2 cos^2 θ (1+tan^2 θ)−(1/2)tan^2 θ = (5/2) (1/2)tan^2 θ= (3/2) ⇒ tan θ = ±(√3) θ = 60° , 120° .

$${At}\:{half}\:{the}\:{max}.\:{height}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} −\:\frac{{mgH}}{\mathrm{2}} \\ $$$$\:{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:...\left({i}\right) \\ $$$${Also}\:{as}\:{given}\:\:\:{u}\mathrm{cos}\:\theta\:=\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:{v} \\ $$$${or}\:\:{v}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\:\:\:\:\:\:\:\:\:\left({ii}\right) \\ $$$${equating}\:\left({i}\right)\:{with}\:\left({ii}\right) \\ $$$$\:{u}^{\mathrm{2}} −\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}\:=\:\frac{\mathrm{5}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{\mathrm{2}} \theta\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{\mathrm{2}} \theta=\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\Rightarrow\:\:\mathrm{tan}\:\theta\:=\:\pm\sqrt{\mathrm{3}}\: \\ $$$$\theta\:=\:\:\mathrm{60}°\:,\:\mathrm{120}°\:. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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