Question Number 73441 by Learner-123 last updated on 12/Nov/19 | ||
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Answered by mr W last updated on 12/Nov/19 | ||
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$$\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}}=\left(\frac{\mathrm{7}+{d}}{\mathrm{7}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{d}=\mathrm{7}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{5}.\mathrm{12}\:{m} \\ $$$$ \\ $$$${v}_{\mathrm{0}} {t}=\mathrm{7} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\Rightarrow{t}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}} \\ $$$${v}_{\mathrm{0}} =\mathrm{7}\sqrt{\mathrm{5}}=\mathrm{15}.\mathrm{65}\:{m}/{s} \\ $$ | ||
Commented by Learner-123 last updated on 12/Nov/19 | ||
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$${how}\:{comes}\:{the}\:\mathrm{1}^{{st}} \:{line}? \\ $$ | ||
Commented by mr W last updated on 12/Nov/19 | ||
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Commented by mr W last updated on 12/Nov/19 | ||
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$${parabola}: \\ $$$${y}={Ax}^{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} ={Ax}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${y}_{\mathrm{2}} ={Ax}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{y}_{\mathrm{2}} }{{y}_{\mathrm{1}} }=\left(\frac{{x}_{\mathrm{2}} }{{x}_{\mathrm{1}} }\right)^{\mathrm{2}} \\ $$ | ||
Commented by Learner-123 last updated on 13/Nov/19 | ||
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$${thanks}\:{sir}. \\ $$ | ||