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-

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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