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The maximum range of a bullet fired from a toy pistol mounted on a car at rest is 40m. What will be acute angle of inclination of pistol for max. Range when the car is moving in the direction f firing with uniform velocity v=20m/s, not the horizontal surface?
We have Range of a projectile = u^2 sin 2theta/g Therefore, u^2 sin 2 theta /g = 40400*sin 2 theta /10 = 40 Hence, theta = 45 degree
u^2/g = 100=> u = 20 m/s t = 2ugsina and R = (ucosa+20)(2usina/g)Put dR/da = 0 => cosa = ½=> a = 60°.
The initial velocity,
when at rest, the range of the projectile is u2 sin(2x) / g. This is maximum when sin(2x)=1.So, 40 = u2 / g => u = 20Now, as the car is moving with 20 m/s, the initial velocity of the projectile in the x direction is (20 + u cos(x)). As acceleration along the x direction is zero, range is T * V=> (2u sin(x) / g)* (20 + u cos(x))on differentiating and equating to zero (to get the maximum) we get x =60 Hope that helps!
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