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Derivation of Position-Time Relation by Graphical Method

A helicopter ...

Question

A helicopter is rising up with speed of 10m/s .A packet is dropped from helicopter when it is 495 m above the ground .Cakculate with what velocity and after how long will the packet hit the ground ? What will be the height of helicopter at that time?

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Solution

Dear Student,

$u = 10 m / s, h = 495 m f r o m s i g n c o n v e n t i o n t a k e u p w a r d a s + v e a n d d o w n w a r d a s - v e u = 10 m / s, h = - 495 m, a = - g l e t t h e v e l o c i t y b e v a t t h e g r o u n d v = - \sqrt{100 + 2 * 10 * 495} = - 100 m / s - v e s h o w d o w n w a r d t i m e t a k e n b e t t = \frac{v - u}{a} = \frac{- 100 - 10}{- 10} = 11 s e c h e l i c o p t e r r i s e f u r h t e r i n 11 s e c H = 495 + 10 * 11 = 495 + 110 = 605 m R e g a r d s$

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