Here's how to calculate a bullet's trajectory, specifically, the time aloft, the point of landing and the highest point in its path. In this example, certain assumptions have been made, for computational simplicity: negligible air resistance, no wind and insufficient firing distance for the Earth's rotation to have effect.
- Skill level:
Things you need
- The initial velocity of the bullet, v(i)
- The angle of the gun, (theta), relative to horizontal
- The height of the target relative to the gun, y(f)
- A calculator that calculates trigonometric functions
First, the shape of the arc must be determined. If the angle is initially downward, then the highest point is already known to be the firing position. Even an upward firing angle might have the target point as the highest point, if the angle is shallow enough or y(f) high enough. This can be determined in step 4 when time aloft is determined.
If the firing angle, theta, is that between the bullet's initial trajectory and horizontal, then the initial vertical velocity is v(i) x sin(theta) .
Time aloft, t, is found using a position equation:
y(f) = v(i) x sin(theta) t - (0.5) g x t^2 , where g = 9.8 meters / sec^2 .
All variables are known except time aloft, t, so t can be solved for, using the quadratic formula:
axe^2 + bx + c = 0 x = [ -b +/- sqrt(b^2-4ac) ] / 2a
If more than one solution for t is allowed, because y(f) is greater than 0, then the first result corresponds to when y=y(f) on the flight path up, and the second to when y=y(f) on the way down.
If y(f) is less than 0, then only one real solution for t was allowed, the other one being negative.
Determining the shape of the arc
If theta is less than 0, then we already know the maximum height is the initial height, y(i)=0.
If there was more than one time, t, at which the bullet reaches y(f), then the smaller t corresponds to a flight path where y(f) is the highest point. The larger t corresponds to the bullet peaking at a higher height y before returning to y(f).
To solve for this peak height, use the formula v(t)=v(0)-9.8t to solve for t when the vertical velocity was zero. In other words, for what time, t, is v(i) x sin(theta) = 9.8t ?
Solving for t and plugging into a height formula gives your maximum height:
y = v(i) x sin(theta) x t - 4.9 t^2
The same approach is used to solve for maximum height if only one solution for t was allowed.
Identifying maximum height
To determine the horizontal distance travelled by the time the bullet reaches height y(f), first calculate the bullet's initial horizontal velocity: v(i) x cos(theta).
Plug the time, t, when the bullet arrives at the final height, y(f), into a position formula that uses the horizontal velocity.
x(f) = v(i) x cos(theta) x t
Assuming no wind resistance, there is no acceleration term on the right-hand side.
If there were more than one time t when y was at y(f), then two horizontal positions x(f) will be valid, with the highest point achieved being y(f) for the smaller of the two x(f).
The final horizontal and vertical positions and the highest point achieved are now known, thus determining the trajectory of the bullet.
- 20 of the funniest online reviews ever
- 14 Biggest lies people tell in online dating sites
- Hilarious things Google thinks you're trying to search for