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Grade: 12
        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?
one year ago

Answers : (4)

Akshat Kumar tripathi
42 Points
							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
						
one year ago
K
13 Points
							
     u^2/g = 100
=> u = 20 m/s
 
t = 2ugsina and R = (ucosa+20)(2usina/g)
Put dR/da = 0 => cosa = ½
=> a = 60°.
one year ago
Deepika Pant
28 Points
							The initial velocity,
						
11 months ago
Anagha
13 Points
							
when at rest, the range of the projectile is usin(2x) / g. This is maximum when sin(2x)=1.
So, 
40 = u/ g => u = 20
Now, 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! 
 
14 days ago
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