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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: projectile...answer with detailed soln...rates assured...
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svj29 (2037)

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1. A ball is thrown from a point on ground at some angle of projection. At the same time a bird starts from a point directly above the point of projection at a height h horizontally with speed u. Given that in its flight ball just touches the bird at one point, find the distance where the ball strikes...
Ans: 2u underroot 2h/g
 
2. Average velocity of a particle is a projectile motion between its starting point and the highest point of its trajectory is:
 
Ans: u/2  underroot 1 + 3 cos(square) theta...

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anchitsaini (4332)

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Since the ball touches the bird at only one point , the following conclusion are drawn -->
1)The bird is touched at the highest point .
2)The horizontal velocity of the ball is u

Let the velocity of ball be v at an angle to the horizontal

thus , v cos = u
or
v = u / cos

h = [v2 sin2 ]/ 2g

   =  [u2 sin2 ]/ [2g cos2 ]

we have to find the range in terms of h and u

we know that
R = v2sin2/g

   =u2sin2/ gcos2

   =2u2sin/ gcos

   = 2u [2 [u2sin2/ 2g2cos2 ] ]

   = 2u [2h/g]

Impossible To be Impossible is Impossible
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adityakaushik_1990 (10)

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average velocity =total displacement / total time taken
(vav )x = (R/2) / (T/2) = u cos(theta)
(vav)y = (H/2) / (T/2) = {u sin(theta) }/2
now vav =underoot of { ( vav x )^2 + (v av y)^2}
     
here H=max height 
and T= total time      
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