# Range Equation Limitation question

## Homework Statement

First off, this is NOT a homework problem. I do not need an actual "answer", but I do have a conceptual question. Here it is:

Consider a projectile motion problem such as a baseball being hit. Assume that it is hit and caught at the same height above the ground. In this instance, I believe that the range equation will NOT work. Why not? Most textbooks claim that the so-called range equation works as long as the launch height and final height are equal. However, in such an instance, the range equation fails to give the correct launch angle.

## Homework Equations

$$R = v_o^2/g * sin (2*\Theta)$$

## The Attempt at a Solution

N/A

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nasu
Gold Member
In what way it will not work? What makes you believe this?

Why shouldn't it work?
The proof and hence the equation does not depend on the height as long as max Height is not so high that their are changes in g.
I think your confusion lies in the fact that the same range can be obtained from two different angles ##\theta \ and \ \pi/2##

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I figured it would work, and am uncertain why it does not, or where I made an error. For example, consider the following: R = 35 m, t = 3.2s, y_o = 1.4 m, v_o = 19 m/s with v_y = 16 and v_x = 11, therefore theta = 55 degrees (by the way these values are from an old exam). The range formula does not properly compute theta, unless I made a silly error in which case I would like it pointed out. Thank you!

You don't need the range formula for finding ##\theta##.
Use the fact ##v_y/v_x=tan(\theta)##
EDT-And the answer should be ##55.49^o##

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You don't need the range formula for finding ##\theta##.
Use the fact ##v_y/v_x=tan(\theta)##

Yes, I know that, thank you. However, that is not my question. My question is: why does the range equation not give you the correct value for theta?

What did you take g as?

gneill
Mentor
Yes, I know that, thank you. However, that is not my question. My question is: why does the range equation not give you the correct value for theta?

It does! But there are two possible angles that satisfy the equation, and your "solving" it returned only one of the two, corresponding to the principal value returned by the sin-1 function. To get the other value, a little trig is involved. Consider the following situation on the unit circle:

https://www.physicsforums.com/attachment.php?attachmentid=62240&stc=1&d=1380292338

Two angles, ##\alpha## and ##\alpha '## will yield the same value for sine. The definition of the range and domain of the inverse sine function means it returns only the angle in the first quadrant. It's up to you to determine what ##\alpha '## is from there, if ##\alpha '## is the angle you need.

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nasu
Gold Member
I figured it would work, and am uncertain why it does not, or where I made an error. For example, consider the following: R = 35 m, t = 3.2s, y_o = 1.4 m, v_o = 19 m/s with v_y = 16 and v_x = 11, therefore theta = 55 degrees (by the way these values are from an old exam). The range formula does not properly compute theta, unless I made a silly error in which case I would like it pointed out. Thank you!

What is the actual text of the problem?
How is R defined here? Does it say that the end point is at the same height as the initial point?
It looks over-determined.

And what Gneill said, for sure.