A-grasshopper-can-jump-a-maximum-horizontal-distance-of-40-cm-If-it-spends-negligible-time-on-the-ground-then-in-this-case-its-speed-along-the-horizontal-road-will-be-

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Question Number 15418 by Tinkutara last updated on 10/Jun/17

$$\mathrm{A}\mathrm{grasshopper}\mathrm{can}\mathrm{jump}\mathrm{a}\mathrm{maximum}$$$$\mathrm{horizontal}\mathrm{distance}\mathrm{of}40\mathrm{cm}.\mathrm{If}\mathrm{it}$$$$\mathrm{spends}\mathrm{negligible}\mathrm{time}\mathrm{on}\mathrm{the}\mathrm{ground}$$$$\mathrm{then}\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{its}\mathrm{speed}\mathrm{along}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{road}\mathrm{will}\mathrm{be}?$$

Answered by mrW1 last updated on 10/Jun/17

$$\mathrm{let}$$$$\mathrm{w}=\mathrm{velocity}\mathrm{of}\mathrm{grasshopper}\mathrm{at}\mathrm{start}$$$$\theta =\mathrm{angle}\mathrm{at}\mathrm{start}$$$$\mathrm{u}=\mathrm{velocity}\mathrm{horizontal}$$$$\mathrm{v}=\mathrm{velocity}\mathrm{vertical}\mathrm{at}\mathrm{start}$$$$\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$$$$$$$$\mathrm{u}=\mathrm{wcos}\theta$$$$\mathrm{v}=\mathrm{wsin}\theta$$$$\mathrm{y}=\mathrm{vt}-{5\mathrm{t}}^2=\frac{{\mathrm{v}}^2}{100}-5({\mathrm{t}}^2-2\times \frac{\mathrm{v}}{10}\mathrm{t}+\frac{{\mathrm{v}}^2}{100})=\frac{{\mathrm{v}}^2}{100}-5{(\mathrm{t}-\frac{\mathrm{v}}{10})}^2$$$$\max .\mathrm{y}\mathrm{at}\mathrm{t}=\frac{\mathrm{v}}{10}$$$$\mathrm{time}\mathrm{for}\mathrm{a}\mathrm{complete}\mathrm{jump}\mathrm{T}=2\times \frac{\mathrm{v}}{10}$$$$\mathrm{x}=\mathrm{uT}$$$$\mathrm{x}=\mathrm{u}\times 2\times \frac{\mathrm{v}}{10}=\mathrm{wcos}\theta \times 2\times \frac{\mathrm{wsin}\theta }{10}$$$$=\frac{{\mathrm{w}}^2}{10}\times \sin 2\theta$$$$\max .\mathrm{x}\mathrm{if}\sin 2\theta =1\mathrm{or}\theta =45°$$$$\max .\mathrm{x}=\frac{{\mathrm{w}}^2}{10}=\mathrm{d}=40\mathrm{cm}$$$$\mathrm{w}=\sqrt{10\mathrm{d}}$$$$\mathrm{u}=\mathrm{wcos}\theta =\sqrt{10\mathrm{d}}\times \frac{1}{\sqrt{2}}=\sqrt{5\mathrm{d}}(=\sqrt{\frac{\mathrm{gd}}{2}})$$$$\mathrm{u}=\sqrt{5\times 0.4}=\sqrt{2}\approx 1.4\mathrm{m}/\mathrm{s}\approx 5\mathrm{km}/\mathrm{h}$$$$$$$$\Rightarrow \mathrm{the}\mathrm{grasshopper}\mathrm{has}\mathrm{a}\mathrm{speed}\mathrm{of}$$$$\mathrm{about}5\mathrm{km}/\mathrm{h}.$$

Commented by Tinkutara last updated on 10/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by ajfour last updated on 10/Jun/17

$$maxhorizontalrange=\frac{u^2}{g}$$$$\frac{u^2}{g}=\frac{2}{5}m,\Rightarrow u=2m/s(withg=10m/s^2)$$$$\theta =\frac{\pi }{4}formax.range$$$$speedalonghorizontalis$$$$u\cos \frac{\pi }{4}=\left(2m/s\right)\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}m/s.$$

Commented by Tinkutara last updated on 10/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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