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Question Number 178582 by Best1 last updated on 19/Oct/22

show that Range of the ff projection obtained by algebric expression R=(((ucosθ)(usinθ)+(√((usinθ)^2 +2gh)))/g) help me please

$${show}\:{that}\:{Range}\:{of}\:{the}\:{ff}\:{projection}\: \\ $$$${obtained}\:{by}\:{algebric}\:{expression} \\ $$$${R}=\frac{\left({ucos}\theta\right)\left({usin}\theta\right)+\sqrt{\left({usin}\theta\right)^{\mathrm{2}} +\mathrm{2}{gh}}}{{g}}\:\:\:{help}\:{me}\:{please} \\ $$

Commented by Best1 last updated on 18/Oct/22

please help me

$${please}\:{help}\:{me} \\ $$

Commented by Ar Brandon last updated on 18/Oct/22

dimension of (sinθ^2 )≠dimension of 2gh=M^2 T^(−2) Check question.

$$\mathrm{dimension}\:\mathrm{of}\:\left(\mathrm{sin}\theta^{\mathrm{2}} \right)\neq\mathrm{dimension}\:\mathrm{of}\:\mathrm{2}{gh}={M}^{\mathrm{2}} {T}^{−\mathrm{2}} \\ $$$$\mathrm{Check}\:\mathrm{question}. \\ $$

Commented by Best1 last updated on 19/Oct/22

i corrected it now help me

$${i}\:{corrected}\:{it}\:{now}\:{help}\:{me} \\ $$

Answered by Ar Brandon last updated on 18/Oct/22

For horizontal displacement u_x =ucosθ ⇒x=u_x t ⇒R=(ucosθ)t ⇒t=(R/(ucosθ)) ...(i) For vertical displacement y=u_y t+(1/2)at^2 ⇒h=(usinθ)t−(1/2)gt^2 ⇒gt^2 −2(usinθ)t+2h=0 ⇒t=((2usinθ±(√(4u^2 sin^2 θ−8gh)))/(2g)) ⇒t=((usinθ+(√(u^2 sin^2 θ−2gh)))/g) ...(ii) (i)=(ii) ⇒(R/(ucosθ))=((usinθ+(√(u^2 sin^2 θ−2gh)))/g) ⇒R=(((usinθ)(ucosθ)+ucosθ(√(u^2 sin^2 θ−2gh)))/g)

$$\mathrm{For}\:\mathrm{horizontal}\:\mathrm{displacement} \\ $$$${u}_{{x}} ={u}\mathrm{cos}\theta\:\Rightarrow{x}={u}_{{x}} {t} \\ $$$$\Rightarrow{R}=\left({u}\mathrm{cos}\theta\right){t}\:\Rightarrow{t}=\frac{{R}}{{u}\mathrm{cos}\theta}\:...\left({i}\right) \\ $$$$\mathrm{For}\:\mathrm{vertical}\:\mathrm{displacement} \\ $$$${y}={u}_{{y}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$$\Rightarrow{h}=\left({u}\mathrm{sin}\theta\right){t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\Rightarrow{gt}^{\mathrm{2}} −\mathrm{2}\left({u}\mathrm{sin}\theta\right){t}+\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}\mathrm{sin}\theta\pm\sqrt{\mathrm{4}{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta−\mathrm{8}{gh}}}{\mathrm{2}{g}} \\ $$$$\Rightarrow{t}=\frac{{u}\mathrm{sin}\theta+\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta−\mathrm{2}{gh}}}{{g}}\:...\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right) \\ $$$$\Rightarrow\frac{{R}}{{u}\mathrm{cos}\theta}=\frac{{u}\mathrm{sin}\theta+\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta−\mathrm{2}{gh}}}{{g}} \\ $$$$\Rightarrow{R}=\frac{\left({u}\mathrm{sin}\theta\right)\left({u}\mathrm{cos}\theta\right)+{u}\mathrm{cos}\theta\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta−\mathrm{2}{gh}}}{{g}} \\ $$

Commented by Best1 last updated on 18/Oct/22

is this the same as (((ucosθ)(usinθ)+(√(sinθ^2 −2gh)))/g)???

$${is}\:{this}\:{the}\:{same}\:{as}\:\frac{\left({ucos}\theta\right)\left({usin}\theta\right)+\sqrt{{sin}\theta^{\mathrm{2}} −\mathrm{2}{gh}}}{{g}}??? \\ $$

Commented by Ar Brandon last updated on 18/Oct/22

I think this isn′t correct as the dimensions aren′t the same.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{isn}'\mathrm{t}\:\mathrm{correct}\:\mathrm{as}\:\mathrm{the}\:\mathrm{dimensions}\: \\ $$$$\mathrm{aren}'\mathrm{t}\:\mathrm{the}\:\mathrm{same}. \\ $$

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