Range Equation Limitation question

• astropi
In summary, the conversation discusses the use of the range equation in projectile motion problems where the launch and final heights are equal. While most textbooks claim that the range equation works in this scenario, it is argued that it fails to give the correct launch angle. The conversation also mentions the use of trigonometry to find the correct angle in this situation.

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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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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Enigman said:
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?

astropi said:
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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astropi said:
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.

1. What is the range equation and its limitations?

The range equation is a mathematical formula used to calculate the maximum distance a projectile can travel given its initial velocity, angle of launch, and gravitational acceleration. Its limitations include assuming a constant gravitational acceleration, neglecting air resistance, and not accounting for factors such as wind and elevation.

2. How accurate is the range equation in real-world situations?

The accuracy of the range equation depends on how well the assumptions hold true in a given situation. In ideal conditions, it can provide a fairly accurate estimate of projectile range. However, in real-world situations where factors like air resistance and wind play a significant role, the actual range may differ significantly from the calculated value.

3. Can the range equation be used for all types of projectiles?

The range equation is derived specifically for projectiles launched at an angle from a horizontal surface. While it can be used for a wide range of projectiles, it may not accurately predict the range for more complex or non-conventional projectiles.

4. How does changing the initial velocity or launch angle affect the range?

According to the range equation, the range is directly proportional to the square of the initial velocity and the sine of the launch angle. This means that increasing the initial velocity or launch angle will result in a longer range, while decreasing them will result in a shorter range.

5. What are some real-life applications of the range equation?

The range equation has various applications in fields such as physics, engineering, and ballistics. It is commonly used in designing and analyzing projectile motion in sports, military operations, and space exploration. It also helps in estimating the maximum range of weapons and in determining the optimal launch angle for distance and accuracy.