A-cyclist-is-travelling-down-a-hill-at-a-speed-of-9-2m-s-The-hillside-makes-an-angle-of-6-3-with-the-horizontal-Calculate-for-the-cyclist-i-the-vertical-speed-ii-horizontal-speed-

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Question Number 31143 by NECx last updated on 03/Mar/18

$$Acyclististravellingdowna$$$$hillataspeedof9.2m/s.The$$$$hillsidemakesanangleof6.3°$$$$withthehorizontal.Calculate,$$$$forthecyclist:$$$$\left(i\right)theverticalspeed$$$$\left(ii\right)horizontalspeed$$

Answered by MJS last updated on 03/Mar/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{a}\mathrm{rectangular}\mathrm{triangle}$$$$a^2+b^2=9.2^2$$$$\alpha =6.3°$$$$\beta =(90-6.3)°=83.7°$$$$\gamma =90°$$$$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=$$$$a=c\times \sin \alpha$$$$b=c\times \sin \beta =c\times \cos \alpha$$$$a=9.2\sin 6.3=1.009555\dots \mathrm{vertical}\mathrm{speed}$$$$b=9.2\cos 6.3=9.144440\dots \mathrm{horizontal}\mathrm{speed}$$

Commented by NECx last updated on 03/Mar/18

$$Thankyousomuch.$$

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