# Physics: Confusion on "range" equation when there is a difference in height

Discussion in 'MCAT Study Question Q&A' started by R35, May 13, 2014.

1. ### R35 2+ Year Member

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May 15, 2012
I'm confused with the range equation. I am using TBR and am unclear on a few concepts:

1) As I understand it, if the initial and final height of a projectile are the same (ex. it is launched on a field and ends on the same field), I can use the equation: r = (vo²Sin2theta)/g .

2) If the heights are different, which equation can I now use? Also, in one of TBR's physics sample problems, I believe they used the above equation for a cannonball falling over a cliff. I thought that wasn't the right equation.

3. ### Chrisz 2+ Year Member

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Feb 18, 2014
Physics depends on the specific details of the problem. Without knowing the specific conditions, such as what values we have at hand, we cant decide in what way we be able solve the question. If we have the specific velocities and specific thetas at these "TWO" different heights, we still can employ the above formula to solve the distance it has traveled horizontally. Actually, if the initial velocity and theta and height difference are known, another way better than this can be used.

Last edited: May 13, 2014
4. ### NextStepTutor_1Next Step Test Prep TutorExhibitor 2+ Year Member

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I personally don't like to use equations for very specific cases because then you have to worry about remembering when to use which equation and if this case fits that equation. Instead, just use the main kinematic equations relating vi, vf, a, d, and t. Based on these equations you should be able to get the correct answer for any case. In this example, if you have different heights, just find the initial vertical velocity, then find the time it takes to get to the top of it's trajectory using: (vf - vi) = at, where a = -10m/s^2, and vf = 0, and vi is whatever they give you. Then you can find the time it takes to get from the top of the trajectory to the distance you desire using the equation: d = 1/2 * a * t^2 + vi*t, with vi = 0 (at top of trajectory), and a = -10, d = whatever is needed, and find t. Based on this, you will get the total time it takes to travel from the start to the finish, and then just multiply by the horizontal velocity to get the horizontal displacement.

Hope that helps!

5. ### syoungMS-3 5+ Year Member

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MDApps:
If you are having two different heights, you can break the problem up into two pieces.

Starting from high and ending low:
Consider breaking this up into two parts
1) Solve the time for projectile to land at the same height again
2) solve the time it would take for the projectile to just fall from the original height down to the low height

Starting from low and ending high:
Reverse the problem and pretend to start from high and ending low

6. ### mehc012Big Damn Hero 5+ Year Member

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Just don't forget that Vy0 in part 2) is the final velocity (or the opposite of the initial vertical velocity) from part 1. I find it easier to split it like this:

1) solve for the max height and time for projectile to reach the max height
2) solve for the time for the projectile to fall from the max height to the low height

7. ### FCMike11 5+ Year Member

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Aug 24, 2011
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Do Voy/g and you get the time to the apex and from there you can calculate the total time in the air. From there multiply your Vox by the total time = the range. Im pretty sure it is Vox, but it may just the Vo. Anyone on that?

8. ### mehc012Big Damn Hero 5+ Year Member

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It would be V0x, but you can't just get the time from/to the apex, because it is not a symmetric parabola

9. ### FCMike11 5+ Year Member

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Initial y velocity divided by gravity will give you time to the apex. After that you know how high you are (Voy x Time to the apex + initial launch height). At that point you have total height and can approximate the time to fall the apex distance with the TBR chart (1s=5m, 2s=20m) so on and so forth. Add that time to your time the apex and you have the total time.

10. ### mehc012Big Damn Hero 5+ Year Member

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Yes, I know that...that is essentially what I put in an above post. However, your last post was missing a few steps.

11. ### catinthehat1234

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Jun 2, 2016
I have a quick question...for the range equation (Vo^2*2sin(theta)/g), does it pertain to x or y component of velocity? Thanks!

12. ### BerkReviewTeachCompany Rep & Bad SingerExhibitor 10+ Year Member

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Neither x nor y; it is the total velocity.