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Question Number 178632 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))

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

Commented by Best1 last updated on 19/Oct/22

please help me

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

Commented by mr W last updated on 19/Oct/22

what you gave is wrong. Ar Brandon  sir has given you the correct answer  and corresponding working.

$${what}\:{you}\:{gave}\:{is}\:{wrong}.\:{Ar}\:{Brandon} \\ $$$${sir}\:{has}\:{given}\:{you}\:{the}\:{correct}\:{answer} \\ $$$${and}\:{corresponding}\:{working}. \\ $$

Commented by Best1 last updated on 19/Oct/22

i corrected it sir

$${i}\:{corrected}\:{it}\:{sir}\: \\ $$

Commented by mr W last updated on 19/Oct/22

the horizontal range is  R=((u cos θ)/g)(u sin θ+(√(u^2  sin^2  θ+2gh)))  with h≥0.

$${the}\:{horizontal}\:{range}\:{is} \\ $$$${R}=\frac{{u}\:\mathrm{cos}\:\theta}{{g}}\left({u}\:\mathrm{sin}\:\theta+\sqrt{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gh}}\right) \\ $$$${with}\:{h}\geqslant\mathrm{0}. \\ $$

Commented by Best1 last updated on 19/Oct/22

please show me the step

$${please}\:{show}\:{me}\:{the}\:{step} \\ $$

Commented by Best1 last updated on 19/Oct/22

$$ \\ $$

Commented by mr W last updated on 19/Oct/22

Ar Brandon sir has shown the steps  in Q178582.

$${Ar}\:{Brandon}\:{sir}\:{has}\:{shown}\:{the}\:{steps} \\ $$$${in}\:{Q}\mathrm{178582}. \\ $$

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