A body is projected at an angle 35° with an initial speed of 45m/s. What is the velocity of the body after 2s and the angle made with the horizontal axis? [g=9.81ms^(−2) ]


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Question Number 31148 by NECx last updated on 03/Mar/18

$A b o d y i s p r o j e c t e d a t a n a n g l e$ $35 ° w i t h a n i n i t i a l s p e e d o f 45 m / s .$ $W h a t i s t h e v e l o c i t y o f t h e b o d y$ $a f t e r 2 s a n d t h e a n g l e m a d e w i t h$ $t h e h o r i z o n t a l a x i s ? [g = 9.81 {m s}^{- 2}]$
Commented by NECx last updated on 03/Mar/18

$f o r t h e v e l o c i t y i n t i m e t, w e h a v e$ $V (t) = u \sin θ - g t$ $v (2) = 45 s i n 35 - 9.81 \times 2$ $= 25.81094 - 19.62$ $= 6.19 m / s$ $t h e n I g o t c o n f u s e d o n s e e i n g a n g l e$ $m a d e w i t h t h e h o r i z o n t a l . P l e a s e$ $h e l p . T h a n k s$
	Answered by mrW2 last updated on 03/Mar/18

	$u = 45 m / s$ $u_{x} = 45 \cos 35 °$ $u_{y} = 45 \sin 35 °$ $v_{x} (t) = u_{x}$ $v_{y} (t) = u_{y} - g t$ $v (t) = \sqrt{v_{x}^{2} + v_{y}^{2}}$ $\tan θ (t) = \frac{v_{y}}{v_{x}}$ $θ (t) = \tan^{- 1} (\frac{v_{y}}{v_{x}})$ $a t t = 2 s :$ $v_{x} (2) = 45 \cos 35 ° = 36.86 m / s$ $v_{y} (2) = 45 \sin 35 ° - 9.81 \times 2 = 6.19 m / s$ $v (2) = \sqrt{36 . 86^{2} + 6 . 19^{2}} = 37.4 m / s$ $θ (2) = \tan^{- 1} \frac{6.19}{36.86} = 9.5 °$
	Commented by NECx last updated on 03/Mar/18

	$t h a n k y o u s o m u c h s i r$
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