# Physics Experiment Question; Projectile Motion

1. Feb 15, 2005

### rumaithya

Hello, I have a question that says' Derive an equation for calculating the theoretical range.

I did like half of it, I got

t = sqrt[ (2y) / a ]

from the kinematic equation ( X = X0 + V0 t + 1/2 a t^2 ). How would I find the theoretical range after this????

And how to prove that the theory is consistent with the experiment?

2. Feb 15, 2005

### Staff: Mentor

Range of what? A projectile with a certain initial speed? Note that the range--the horizontal distance traveled--will depend on the angle that the projectile is launched.

You'll need to combine equations for both horizontal and vertical motion.

3. Feb 15, 2005

### Jameson

If you combine the two equations for projectile motion

$$y = y_0 + v_it + .5at^2$$
$$x = x_0 + v_it +.5at^2$$

into one equation, the highest range would be the vertex of the parabola, which is half of the total time I believe. I would suggest solving for t, then divinding that by 2 to get the time when the highest point is reached, which is what I'm assuming the "range" means.

4. Feb 15, 2005

### Staff: Mentor

(1) Only the vertical component of the motion is accelerated.
(2) "Range" generally means horizontal distance traveled. (If you shot the projectile straight up, it would fall straight down. Range = 0.)

But, yes, the highest point attained is the vertex of a parabola.

5. Feb 15, 2005

### j_confused

You'd need to break your theoretical initial velocity and angle of projection down into the initial X and Y components.
(R being initial velocity magnitude, O being initial angle)

Vox R x cosO
Voy R x sinO

Then, you'll also have to break down your equations for motion into X and Y, I'm going to assume by range you mean distance which is X-motion...
(Xo being initial X position -usually 0-; Vox being the x-component's initial velocity)

X X(t) = Xo + Voxt + 0.5at^2

(Remember that the acceleration for the x-component is ZERO!)
Now, since you have a theoretical Vox, you can plug it into the above eq.

*edit* You'd also have to do the same for the Y component, and derive that equation. Then set that equal to zero, because that will tell you the two time the projectile will cross the X-axis. Pick the positive time and substitute it into your position-time function for X.

I believe that the above is right as for range, now as for the experimental proving bit, that I can't help with.

Last edited: Feb 15, 2005