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

$A c y c l i s t i s t r a v e l l i n g d o w n a$ $h i l l a t a s p e e d o f 9.2 m / s . T h e$ $h i l l s i d e m a k e s a n a n g l e o f 6.3 °$ $w i t h t h e h o r i z o n t a l . C a l c u l a t e,$ $f o r t h e c y c l i s t :$ $(i) t h e v e r t i c a l s p e e d$ $(i i) h o r i z o n t a l s p e e d$
	Answered by MJS last updated on 03/Mar/18

	${it}^{'} s just a rectangular triangle$ $a^{2} + b^{2} = 9 . 2^{2}$ $α = 6.3 °$ $β = (90 - 6.3) ° = 83.7 °$ $γ = 90 °$ $\frac{a}{\sin α} = \frac{b}{\sin β} =$ $a = c \times \sin α$ $b = c \times \sin β = c \times \cos α$ $a = 9 . 2 \sin 6.3 = 1.009 555 . . . vertical speed$ $b = 9 . 2 \cos 6.3 = 9.144 440 . . . horizontal speed$
	Commented by NECx last updated on 03/Mar/18

	$T h a n k y o u s o m u c h .$
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