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 the air resistance.
Doubt by Niyati
Solution :
θ = 30°
Rangae (R)= 3 km = 3000 m
We know,
Range of Angular Projectile is given by
R = u2sin2θ/2g
Rangae (R)= 3 km = 3000 m
We know,
Range of Angular Projectile is given by
R = u2sin2θ/2g
R = u2sin2(30°)/2g
R = u2sin(60°)/2g
R = u2(√3/2)/2g
R = u2√3/4g
4Rg/√3 = u2
u2 = 4Rg/√3 — (1)
We know,
For Maximum Range
Rmax = u2/2g
Rmax = [4Rg/√3]/2g [From eq (1)]
Rmax = 2R/√3
Rmax = (2×3)/√3 km
Rmax = 6/√3 km
Rmax = 2√3 km
Rmax = 2×1.732
Rmax = 3.464 km
Rmax = 3.46 km
R = u2sin(60°)/2g
R = u2(√3/2)/2g
R = u2√3/4g
4Rg/√3 = u2
u2 = 4Rg/√3 — (1)
We know,
For Maximum Range
Rmax = u2/2g
Rmax = [4Rg/√3]/2g [From eq (1)]
Rmax = 2R/√3
Rmax = (2×3)/√3 km
Rmax = 6/√3 km
Rmax = 2√3 km
Rmax = 2×1.732
Rmax = 3.464 km
Rmax = 3.46 km
Since the maximum range of the projectile is just 3.46 km, so we can't hit the target which is 5 km away.