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Sir plz make me understand that why are we adding the velocity of car and gun's bullet...does it not change range?..how's that being applied  here I can't understand,thanks

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Asked by vishakhachandan026 10th June 2019, 10:34 AM
Answered by Expert
Answer:
When car is at rest :-
 
Range for projectile motion = u2sin(2α)/(2g)  ; for maximum range, α = 45° and maximum range is given by Rmax = u2/(2g)
Where u is projection velocity and α is angle of projection
 
since maximu range is 40 m,  projection velocity is obtained as  u2 = Rmax (2g)  = 40×2×10   hence u = 20√2  m/s
---------------------
 
when car is moving at velocity 20 m/s in firing direction :-
 
since bullet is fired from moving car its resultant velocity is vector sum of velocity of car and its projection velocity.
 
Let us assume direction of car's movement is along +ve x-axis,  hence velocity of car = 20 i m/s   ................(1)
 
where i is positive unit vector along x-axis.
 
To get maximum range , let us assume bullet is fired so that bullet projection velocity is  ( a i + b j ) m/s
 
where j is positive unit vector along y-axis.
 
Resultant velocity v is vector sum that is given by,  v = (20+a) i + b j
 
we know that for maximum range,  projection angle is 45°, hence  b/(20+a) = tan 45 = 1  or b = 20+a ................(2)
 
since magnitude of projection velocity is  20√2 m/s,  we have   a2 + b2 = (20√2)2 = 800  ..................(3)
 
By substituting for b using eqn.(2) in eqn.(3), we get  a2 + (20+a)2 = 800  or  a2+20a-400 = 0 .................(4)
 
Realistic solution of a of eqn.(4) is obtained as a = 10(√3 - 1)................(5)
 
using eqn(2) and eqn.(5), we get b = 10(√3+1) 
 
Hence projection angle to fire the bullet = tan-1(b/a) = tan-1 [ (√3+1) / (√3-1) ] = 75°
Answered by Expert 10th June 2019, 1:48 PM
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