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.

**Solution:**

Given that θ = 30⁰, R = 3 Km = 3000m

The horizontal range of the fired bullet is given by , R = u^{2} sin 2 θ/g

Or 3000m =(u^{2} /g ) sin(2 x 30⁰) =( u^{2} /g )sin 60⁰ = (u^{2}/g) x √3/2

So, u^{2}/g= 6000/√3 m

**The range of fired bullet is maximum for θ = 45⁰, Hence maximum range is:**

R _{m }= u^{2}/g sin (2 x 45⁰) = (u^{2} / g) sin 90⁰ = 6000/√3 = 3464 m

Which is much smaller than 5 Km. Hence we cannot hit the target at 5 Km.