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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 | 10 Jun, 2019, 10:34: AM
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
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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 Thiyagarajan K | 10 Jun, 2019, 01:48: PM
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