Question

The maximum range of a bullet fired from a toy pistol mounted on a car at rest is 40m. What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with a uniform velocity of 20m/s on a horizontal surface?

Answer 1

Given:-

Maximum Range R₀ = 40m.

Car velocity v =  20m/s.

g = 10m/sec

To Find:-

Acute angle of inclination of the pistol of maximum range.

Solution:-

According to question maximum range is given by R₀ = 40m/sec

Then we know that the maximum range is given by a formula $\frac{u^{2}sin2\theta{} }{g}$ at angle 45°.

Therefor R₀ =  $\frac{u^{2}sin2\theta{} }{g}$  =  40.    (θ = 45)  

           By solving we get; u = 20m/sec.

Now according to horizontal and vertical component we have

Vertical component = 20sinθ.

Horizontal component = 20 + 20cosθ.

Then the time of flight will be = $\frac{2usin\theta{}}{g}$  ( g = 10 and u = 1=) then;

time of flight = $\frac{2x20sin\theta{}}{10}$ = 4sinθ.

Then the horizontal Displacement given by;

R =  ( Horizontal component) x time of flight

R =  ( 20 + 20cosθ) x time of flight

R = ( 20 + 20cosθ)x 4sinθ

R = 80sinθ + 80sinθcosθ

R = 80sinθ + 40sin2θ.

To find the value of θ , differentiate above equation with respect to θ and equating to zero  then we have,

$\frac{dR}{d\theta}$ = $\frac{d}{d\theta}$ ( 80sinθ + 40sin2θ) = 0

Differentiate above and equate to zero, then we get

80cosθ + 40cos2θx2 = 0

cosθ = - cos2θ

cosθ = cos( π - 2θ)

θ = π - 2θ

θ = π/3  = $\frac{\pi }{3}$ = 60°.

Therefore the acute angle of inclination will be 60°