Question

Question 4.29 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 the fixed, and neglect air resistance.

Chapter Motion In A Plane Page 88

Answer 1

We know,
Range = u²sin2∅/g
Range is maximum at ∅ = 45°
e.g maximum range = u²/g

A/C to question ,
Range = u²sin2×30°/g
3000 m = u²×√3/2/g
20000√3 = u² -------(1)
Put in maximum range
Maximum range = (20000√3)/10
= 2000√3 = 2√3 km = 3.464 km

Hence, bullet can't be fired up to 5km { due to maximum range = 3.464 km } with the same speed .

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.