Question-121180

Question Number 121180 by TITA last updated on 05/Nov/20

Commented by TITA last updated on 05/Nov/20

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

Answered by TANMAY PANACEA last updated on 05/Nov/20

y=0.7x−0.02x^2 y=0 0.7x−0.02x^2 =0 x(0.7−0.02x)=0 The stone hit ata distance=((0.7)/(0.02))=((7×100)/(10×2))=35 meter from point O (vcosθ)t=x y=(vsinθ)t−(1/2)gt^2 y=(((vsinθ)/(vcosθ)))x−(1/2)g×(x^2 /(v^2 cos^2 θ)) y=(tanθ)x−((1/2)×g×(1/(v^2 cos^2 θ)))x^2 comparing tanθ=0.7→θ=tan^(−1) (0.7) (1/2)×10×(1/v^2 )×sec^2 θ=0.02 v^2 =((5(1+0.49))/(0.02))=((1.49×5)/(0.02)) v=(√((5×1.49)/(0.02))) 0^2 =(vsinθ)^2 −2gH H=((v^2 sin^2 θ)/(2g)) H=((1.49×5)/(0.02×2×10))×(((0.7)/( (√(1+0.49)))))^2 H=((5×0.49)/(0.02×20))

$${y}=\mathrm{0}.\mathrm{7}{x}−\mathrm{0}.\mathrm{02}{x}^{\mathrm{2}} \\ $$$${y}=\mathrm{0} \\ $$$$\mathrm{0}.\mathrm{7}{x}−\mathrm{0}.\mathrm{02}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}\left(\mathrm{0}.\mathrm{7}−\mathrm{0}.\mathrm{02}{x}\right)=\mathrm{0} \\ $$$${The}\:{stone}\:{hit}\:{ata}\:{distance}=\frac{\mathrm{0}.\mathrm{7}}{\mathrm{0}.\mathrm{02}}=\frac{\mathrm{7}×\mathrm{100}}{\mathrm{10}×\mathrm{2}}=\mathrm{35}\:{meter}\:{from} \\ $$$${point}\:{O} \\ $$$$\left({vcos}\theta\right){t}={x} \\ $$$${y}=\left({vsin}\theta\right){t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${y}=\left(\frac{{vsin}\theta}{{vcos}\theta}\right){x}−\frac{\mathrm{1}}{\mathrm{2}}{g}×\frac{{x}^{\mathrm{2}} }{{v}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta} \\ $$$${y}=\left({tan}\theta\right){x}−\left(\frac{\mathrm{1}}{\mathrm{2}}×{g}×\frac{\mathrm{1}}{{v}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}\right){x}^{\mathrm{2}} \\ $$$${comparing} \\ $$$${tan}\theta=\mathrm{0}.\mathrm{7}\rightarrow\theta={tan}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{7}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\frac{\mathrm{1}}{{v}^{\mathrm{2}} }×{sec}^{\mathrm{2}} \theta=\mathrm{0}.\mathrm{02} \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{5}\left(\mathrm{1}+\mathrm{0}.\mathrm{49}\right)}{\mathrm{0}.\mathrm{02}}=\frac{\mathrm{1}.\mathrm{49}×\mathrm{5}}{\mathrm{0}.\mathrm{02}} \\ $$$${v}=\sqrt{\frac{\mathrm{5}×\mathrm{1}.\mathrm{49}}{\mathrm{0}.\mathrm{02}}}\: \\ $$$$\mathrm{0}^{\mathrm{2}} =\left({vsin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gH}\:\:\: \\ $$$${H}=\frac{{v}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\mathrm{2}{g}} \\ $$$${H}=\frac{\mathrm{1}.\mathrm{49}×\mathrm{5}}{\mathrm{0}.\mathrm{02}×\mathrm{2}×\mathrm{10}}×\left(\frac{\mathrm{0}.\mathrm{7}}{\:\sqrt{\mathrm{1}+\mathrm{0}.\mathrm{49}}}\right)^{\mathrm{2}} \\ $$$${H}=\frac{\mathrm{5}×\mathrm{0}.\mathrm{49}}{\mathrm{0}.\mathrm{02}×\mathrm{20}} \\ $$

Commented by TITA last updated on 05/Nov/20

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

Commented by TANMAY PANACEA last updated on 06/Nov/20

$${most}\:{welcome} \\ $$

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