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Question-103421




Question Number 103421 by aurpeyz last updated on 15/Jul/20
Answered by Worm_Tail last updated on 15/Jul/20
v_y ^2 =(usinθ)^2 −2gs_y ⇒3.7^2 =(usinθ)^2 −(2×10×7.5)  13.69+150=(usinθ)^2 ⇒(usinθ)^2 =163.69  h_(max) =(((usinθ)^2 )/(2g))=((163.69)/(20))=8.1845m  T=((2(usinθ))/g)=((2×(√(163.69)))/(10))=2.5588s  R=v_x T=10.4×2.5588=26.61m
$${v}_{{y}} ^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gs}_{{y}} \Rightarrow\mathrm{3}.\mathrm{7}^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\left(\mathrm{2}×\mathrm{10}×\mathrm{7}.\mathrm{5}\right) \\ $$$$\mathrm{13}.\mathrm{69}+\mathrm{150}=\left({usin}\theta\right)^{\mathrm{2}} \Rightarrow\left({usin}\theta\right)^{\mathrm{2}} =\mathrm{163}.\mathrm{69} \\ $$$${h}_{{max}} =\frac{\left({usin}\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\mathrm{163}.\mathrm{69}}{\mathrm{20}}=\mathrm{8}.\mathrm{1845}{m} \\ $$$${T}=\frac{\mathrm{2}\left({usin}\theta\right)}{{g}}=\frac{\mathrm{2}×\sqrt{\mathrm{163}.\mathrm{69}}}{\mathrm{10}}=\mathrm{2}.\mathrm{5588}{s} \\ $$$${R}={v}_{{x}} {T}=\mathrm{10}.\mathrm{4}×\mathrm{2}.\mathrm{5588}=\mathrm{26}.\mathrm{61}{m} \\ $$
Commented by aurpeyz last updated on 15/Jul/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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