In this post we will understand the property of fired bullet with an example. We will also understand the maximum height, distance and angle of a fired bullet required to the accurate targeting. According to Physics laws we can easily understand this.
Example to understand the property of fired bullet
When a fired bullet at 30⁰ with the horizontal and hits the ground 3 Km away. By adjusting its angle of projection, can we hit a target 5 Km away? Assume the muzzle speed of the bullet to remain unchanged and friction of air may be neglected.
Given that θ = 30⁰, R = 3 Km = 3000m
The horizontal range of the fired bullet is given by , R = u2 sin 2 θ/g
Or 3000m =(u2 /g ) sin(2 x 30⁰) =( u2 /g )sin 60⁰ = (u2/g) x √3/2
So, u2/g= 6000/√3 m
The range of fired bullet is maximum for θ = 45⁰, Hence maximum range is:
R m = u2/g sin (2 x 45⁰) = (u2 / g) sin 90⁰ = 6000/√3 = 3464 m
Which is much smaller than 5 Km. Hence we cannot hit the target at 5 Km.