Question

A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Answer 1

Horizontal range , R = 3km = 3000m

a/c to question, muzzle speed to be fixed.

Let muzzle speed is u m/s

Use formula of horizontal range,

R = u²sin2(30°)/g

3000 = u²sin60°/10

30000/(√3/2) = u²

u² = 60000/√3 = 20000√3m .....(1)

Now, horizontal range , R= u²sin2$\theta$/g

Maximum value of sin2[\tex]\theta[/tex] = 1

So, maximum height can be possible , H =20000√3/10

= 2000√3

= 3464m = 3.464km

So the bullet can maximum hit a target 3.46 Km away. Hence, it's not possible to hit a target 5km away.

Answer 2

Explanation:

Range, R = 3 km

Angle of projection, θ = 30°

Acceleration due to gravity, g = 9.8 m/s2

Horizontal range for the projection velocity u0, is given by the relation:

R = u02 Sin 2θ / g

3 = u02 Sin 600 / g

u02 / g = 2√3 .......(i)

The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,

Rmax = u02 / g ....(ii)

On comparing equations (i) and (ii), we get:

Rmax = 3√3

= 2 X 1.732 = 3.46 km

Hence, the bullet will not hit a target 5 km away.