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 the fixed, and neglect air resistance.
Solution
Given,
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
i.e., 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.
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
i.e., 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.
