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Question Number 18077 by virus last updated on 15/Jul/17

A mango in a tree is located (30,40) from the point of projection of stone.Find the minimum speed and the angle of projevtion of the stone so as to hit the mango

$${A}\:{mango}\:{in}\:{a}\:{tree}\:{is}\:{located}\:\left(\mathrm{30},\mathrm{40}\right)\:{from}\:{the} \\ $$$${point}\:{of}\:{projection}\:{of}\:{stone}.{Find}\:{the}\: \\ $$$${minimum}\:{speed}\:{and}\:{the}\:{angle}\:{of}\:{projevtion} \\ $$$${of}\:{the}\:{stone}\:{so}\:{as}\:{to}\:{hit}\:{the}\:{mango} \\ $$

Answered by ajfour last updated on 15/Jul/17

Equation of trajectory is y=xtan θ−((gx^2 )/(2u^2 ))(1+tan^2 θ) point M(30,40) must lie on it; so 40=30tan θ−((10(900))/(2u^2 ))(1+tan^2 θ) differentiating and equating to zero: 0=30sec^2 θ−((9000)/(2u^2 ))(2tan θsec^2 θ) ⇒ tan θ=((30u^2 )/(9000)) = (u^2 /(300)) ⇒ 40=30((u^2 /(300)))−((9000)/(2u^2 ))(1+(u^4 /(90000))) ⇒ 40=(u^2 /(10))−((9000)/(2u^2 ))−(u^2 /(20)) 40=(u^2 /(20))−((9000)/(2u^2 )) (..×20u^2 gives) 800u^2 =u^4 −90000 (u^2 −400)^2 =90000+160000 u^2 −400=(√(25×10000_ )) u^2 =900 ⇒ u_(min) =30m/s , θ=tan^(−1) ((u^2 /(300)))=tan^(−1) 3 .

$$\:\mathrm{Equation}\:\mathrm{of}\:\mathrm{trajectory}\:\mathrm{is} \\ $$$$\:\mathrm{y}=\mathrm{xtan}\:\theta−\frac{\mathrm{gx}^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{point}\:\mathrm{M}\left(\mathrm{30},\mathrm{40}\right)\:\mathrm{must}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{it};\:\mathrm{so} \\ $$$$\:\mathrm{40}=\mathrm{30tan}\:\theta−\frac{\mathrm{10}\left(\mathrm{900}\right)}{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{differentiating}\:\mathrm{and}\:\mathrm{equating}\:\mathrm{to}\:\mathrm{zero}: \\ $$$$\:\:\mathrm{0}=\mathrm{30sec}\:^{\mathrm{2}} \theta−\frac{\mathrm{9000}}{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{2tan}\:\theta\mathrm{sec}\:^{\mathrm{2}} \theta\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta=\frac{\mathrm{30u}^{\mathrm{2}} }{\mathrm{9000}}\:=\:\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{300}} \\ $$$$\Rightarrow\:\:\mathrm{40}=\mathrm{30}\left(\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{300}}\right)−\frac{\mathrm{9000}}{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{u}^{\mathrm{4}} }{\mathrm{90000}}\right) \\ $$$$\Rightarrow\:\:\mathrm{40}=\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{10}}−\frac{\mathrm{9000}}{\mathrm{2u}^{\mathrm{2}} }−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{20}} \\ $$$$\:\mathrm{40}=\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{20}}−\frac{\mathrm{9000}}{\mathrm{2u}^{\mathrm{2}} }\:\:\:\:\:\:\left(..×\mathrm{20u}^{\mathrm{2}} \:\mathrm{gives}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{800u}^{\mathrm{2}} =\mathrm{u}^{\mathrm{4}} −\mathrm{90000} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{u}^{\mathrm{2}} −\mathrm{400}\right)^{\mathrm{2}} =\mathrm{90000}+\mathrm{160000} \\ $$$$\:\:\:\:\:\mathrm{u}^{\mathrm{2}} −\mathrm{400}=\sqrt{\mathrm{25}×\mathrm{10000}_{} } \\ $$$$\:\:\:\:\:\mathrm{u}^{\mathrm{2}} =\mathrm{900}\:\:\:\:\Rightarrow\:\:\:\:\mathrm{u}_{\mathrm{min}} =\mathrm{30m}/\mathrm{s}\:, \\ $$$$\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{300}}\right)=\mathrm{tan}^{−\mathrm{1}} \mathrm{3}\:. \\ $$

Commented by virus last updated on 15/Jul/17

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by ajfour last updated on 15/Jul/17

Is answer correct, virus?

$$\mathrm{Is}\:\mathrm{answer}\:\mathrm{correct},\:\mathrm{virus}? \\ $$

Commented by virus last updated on 15/Jul/17

yes sir

$${yes}\:{sir} \\ $$

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