Finding range of a projectile from a graph

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    Graph Projectile Range
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Discussion Overview

The discussion revolves around investigating the effect of initial velocity on the range of a projectile, specifically in the context of a report involving a slingshot. Participants explore the relationship between initial velocity and range, and the use of graphs to illustrate this relationship.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks graphs to help illustrate the relationship between initial velocity and range for a projectile.
  • Another participant states that the range of a projectile is proportional to the square of the initial velocity when the angle of projection is constant.
  • A participant clarifies that the report involves a slingshot, where the rubber band is extended to different lengths, affecting the initial velocity while keeping the angle constant.
  • There is a discussion about the meaning of the "gradient" of the graph, with one participant confirming that they meant the slope.
  • One participant suggests that if the formula for range is known, the slope can be derived and related to the initial velocity.
  • Another participant questions the meaningfulness of the condition related to the slope and range.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between the slope of the graph and the range, with some clarifications made but no consensus reached on the best approach to represent the data graphically.

Contextual Notes

There are unresolved aspects regarding the specific graphs that could be used and the implications of the slope in relation to the range of the projectile.

James Adfey
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Hey,
I am writing a report where I am trying to investigate the effect of initial velocity on the range of a projectile. Does anyone know any graphs I could use to help me find the range?
Thanks,
 
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upload_2016-6-18_13-26-32.png


Range of the Projectile fired at an angle θ from the ground with an initial velocity Vi is given by

upload_2016-6-18_13-26-58.png


So Range is proportional to square of initial velocity keeping angle of projection θ as constant.

But I don't understand your question. Which graph you are talking about?
 
Prateek Kumar Jain said:
View attachment 102175

Range of the Projectile fired at an angle θ from the ground with an initial velocity Vi is given by

View attachment 102176

So Range is proportional to square of initial velocity keeping angle of projection θ as constant.

But I don't understand your question. Which graph you are talking about?

Apologies, so the report is being done with a sling shot, where the rubberband will be extended to different lengths, as this will affect the initial velocity. Angle is being kept constant and I am not sure which graph to use, however I am expected to manipulate a graph to show the relationship between initial velocity and range, and was wondering if it was possible to find the range through the gradient of a graph.
 
James Adfey said:
and was wondering if it was possible to find the range through the gradient of a graph
What do you mean of the "gradient" of the graph? Your graph must mean the range ##R## versus the initial velocity ##v,## and in this single-variable function, the gradient is just the slope.
 
tommyxu3 said:
What do you mean of the "gradient" of the graph? Your graph must mean the range ##R## versus the initial velocity ##v,## and in this single-variable function, the gradient is just the slope.
ok yeah I just meant the slope when I said gradient, and was wondering if there were any graphs where the slope is equal to the range, but if not I will just do a graph that shows the range vs initial velocity,
Thanks,
 
James Adfey said:
the slope is equal to the range
If you knew the result ##R=\frac{v^2\sin 2\theta}{g},## then you can easily get the slope ##R'(v)=\frac{2v\sin 2\theta}{g}## and can get what point meets your demand, which, however, the condition seems not meaningful... or?
 
ok ye
tommyxu3 said:
If you knew the result ##R=\frac{v^2\sin 2\theta}{g},## then you can easily get the slope ##R'(v)=\frac{2v\sin 2\theta}{g}## and can get what point meets your demand, which, however, the condition seems not meaningful... or?
ok yeah that makes sense, thanks
 

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