What angle yields the maximum horizontal range for a projectile?

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Homework Help Overview

The discussion revolves around determining the angle that yields the maximum horizontal range for a projectile, specifically in the context of a plastic dart gun. The original poster describes their attempts to calculate the maximum range after firing the dart straight up and mentions having calculated the initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the angle of launch and the horizontal range, with one participant suggesting that a 45-degree angle is optimal for maximum distance. The original poster expresses confusion about how to proceed after calculating the initial velocity and mentions the need to incorporate vectors.

Discussion Status

Some guidance has been provided regarding the optimal angle for maximum range, and the original poster acknowledges the concept of 45 degrees as a known solution. There is an exploration of reasoning behind this angle, including intuitive and rigorous approaches to derive it mathematically.

Contextual Notes

The original poster's calculations are based on the assumption that the total vertical displacement is zero, and they are seeking clarification on how to proceed with their analysis. There is also a mention of a sample problem that the original poster overlooked, indicating potential gaps in their understanding of the topic.

TickleMeElma
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You buy a plastic dart gun, and you wnt to find its max. horizontal range. you shoot the gun straight up, and it takes 4s for the dart to land back at the barrel. What is the maximum horizontal range of your gun?

I got the initial velocity, which is 19.6 m/s. I know I have to incorporate vectors. The initial velocity in the horizontal direction is 19.6m/s (cos x) and in the vertical direction it is 19.6m/s (sin x). Assuming x is the magnitude of the angle. I know that the total displacement in the vertical direction is 0.

And now I am stuck...How do I proceed from here? Please help!

Thanks so very much.
 
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Hint: What value for the angle x will produce the maximum range?
 
Last edited:
Your [itex]\vec{v_i}[/itex] vector points at whatever direction you fire at. To fire for maximum distance you want to fire at a 45 degree angle to the horizontal, which means you'll have velocity vector [itex]\vec{v_i} = 19.6m/s @ 45 \ deg [/tex]. Find how far that goes.[/itex]
 
Thank you so much for your help - yeah, somehow, I missed the sample problem that dealt with the apparently widely-known concept of 45 degrees being the angle that yields the max horizontal replacement. Went back and reworked it with your suggestions and those of the book and got ~39m. :)
 
apparently widely-known concept of 45 degrees being the angle that yields the max horizontal replacement

It can be sort of intuitive. If the angle is 90 (i.e. straight up), the range is zero. If the angle is zero, the range is zero. Thus, by reasoning, the max would be 1/2 way between the two, or 45 degrees.

Actually, there is a rigorous way to prove it using the equations of motions for a projectile y(t) and x(t). By setting y = 0 and solving for t you get the time it takes to complete the motion. Using this t in the equation for x, you get what's called the range equation, which gives you x as a function of angle. Since you want max x, differentiate x with respect to theta, set it equal to zero, and solve for theta.
 

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