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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

Asked by vishakhachandan026 10th June 2019, 10:34 AM
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
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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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