# Qualitative Explanation of Range

1. Apr 7, 2014

### tmobilerocks

1. The problem statement, all variables and given/known data
Explain qualitatively how it is possible for the projectiles to have the same range despite having different launch angles

2. Relevant equations
x = voxt

3. The attempt at a solution
A projectile's range is determined by its x component of the initial velocity and the airtime. A projectile can either have a large x-component of its initial velocity, and a short airtime, or a small x-component of its initial velocity and a long airtime. Therefore, they will have the same range.

Is there a better explanation than this?

2. Apr 7, 2014

### electricspit

Qualitatively:

You have to imagine the fact that in let's say a 2-dimensional plane where the projectile flies, there are two perpendicular directions. Because of this fact, the y-component of velocity is completely independent of the x-component. It only controls motion in the vertical direction; try thinking of each component as if the other were set to zero.

For example, if $v_{ox}$ were set to 0, the projectile would go up, then fall right back down. Now if you slowly increase $v_{ox}$ you will basically be spreading out the rise and fall over more space. The rise and fall will still take exactly the same time as if there were no x-component. In the case where there is an x-component it just will not land where it started!

Mathematically:

The same initial x-component of velocity will lead to the same range, because they correspond to different angles. For example let's say you have two different projectiles with the same initial x-velocity:

$v_1=[\hat{x}+\hat{y}] m/s$

$v_2=[\hat{x}+\sqrt{3}\hat{y}] m/s$

If you haven't been introduced to unit vectors, all these $\hat{x},\hat{y}$ symbols mean basically are x-component and y-component respectively.

So for example with $v_1$ projectile the initial velocity is $1 m/s$ in the x-direction, and $1 m/s$ in the y-direction.

The $v_2$ projectile has an initial velocity of $1 m/s$ in the x-direction and $\sqrt{3} m/s$ in the y-direction.

So the second projectile has much more speed in the y-direction than the first projectile. The angle of both can be found from simple trigonometry:

$\theta_1 = \arctan{1} = \frac{\pi}{4} = 45^\circ$

$\theta_2 = \arctan{\sqrt{3}} = \frac{\pi}{3} = 60^\circ$

So visually you should picture the parabolic path of the projectile, and imagine the two roots of the parabola being it's initial position on the ground, and it's final position. The $45^\circ$ velocity will be much lower to the ground during its flight but will have the same roots as the parabola that had an initial angle $60^\circ$. All that changed was how high it went, which was determined by the differing y-component. The x-component just changed how much over it went, which was in this case exactly the same!

I hope this was clear!