An-object-is-projected-from-a-height-of-80m-above-the-ground-with-a-velocity-of-40m-s-at-an-angle-of-30-degree-to-the-horizontal-What-is-the-tume-of-flight-

Question Number 46629 by Necxx last updated on 29/Oct/18

A n o b j e c t i s p r o j e c t e d f r o m a

h e i g h t o f 80 m a b o v e t h e g r o u n d

w i t h a v e l o c i t y o f 40 m / s a t a n

a n g l e o f 30 d e g r e e t o t h e

h o r i z o n t a l . W h a t i s t h e t u m e o f

f l i g h t ?

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

- h = (u s i n θ) t - \frac{1}{2} {g t}^{2} h e r e d i s p l a c e m e n t = - 80

g r a v i t a t i o n a l a c c e l a r a t i o n d o w n w a r d s o - v e

- 80 = 40 \times \frac{1}{2} \times t - 5 t^{2}

- 80 = 20 t - 5 t^{2}

- 16 = 4 t - t^{2}

t^{2} - 4 t - 16 = 0

t = \frac{4 \pm \sqrt{16 + 64}}{2}

t = \frac{4 \pm 4 \sqrt{5}}{2}

s o t h e a n s w e r i s t = \frac{4 + 4 \sqrt{5}}{2} = 2 + 2 \sqrt{5} s e c o n d

p l s c h e c k

t = 2 \sqrt{5} + 2 = 6.47 s e c

i f w e t a k e g = 9.8

t h e n \dots e q n i s

- 80 = \frac{1}{2} \times 40 t - \frac{1}{2} \times 9.8 t^{2}

9.8 t^{2} - 40 t - 160 = 0

t = \frac{40 \pm \sqrt{1600 + 4 \times 9.8 \times 160}}{2 \times 9.8}

= \frac{40 \pm \sqrt{1600 + 64 \times 98}}{2 \times 9.8}

= \frac{40 \pm \sqrt{7872}}{2 \times 9.8} = \frac{40 \pm 88.72}{2 \times 9.8}

t = \frac{128.72}{2 \times 9.8} = 6.57

Commented by Necxx last updated on 29/Oct/18

t h e o p t i o n s a r e a) 2 b) 4 c) 8 d) 10

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

Commented by Necxx last updated on 29/Oct/18

t h a n k y o u s o m u c h . t h i s w a s a l s o

w h a t i g o t b u t i w a s c o n f u s e d a t

n o t s e e i n g a n y r e l a t e d o p t i o n .

I b e l i e v e y o u p r o f .

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