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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?`
one year ago

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 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!
```
14 days ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions