A Projectile Fired At A 45 Degree Angle

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Homework Help Overview

The problem involves a projectile fired at a 45-degree angle, which must just clear a 6-meter high fence located 100 meters away. Participants are tasked with determining the projectile's initial velocity, focusing on the components of motion and the equations of kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to break down the problem into x and y components and express the horizontal and vertical components of velocity in terms of the initial velocity. There are questions about the use of kinematic equations and the implications of the given parameters.

Discussion Status

Some participants have suggested writing down known variables and equations, while others have explored the relationships between the components of motion. There is an ongoing exploration of how to apply simultaneous equations to find the initial velocity, with various interpretations of the problem being discussed.

Contextual Notes

Participants note the challenge of working backwards from the typical projectile motion problems, questioning assumptions about the launch height and the effects of gravity on the projectile's trajectory.

  • #31
6=9.8t^2*(100/0.7071t)*0.7071

we just told you that you should have a plus in there, not a multiplication!

you do something wrong when you insert the substitution.
 
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  • #32
Ah, found the problem,

s = ut+(1/2) at^2

I used this formula and I got the right answer... thanks anyway guys
 
  • #33
well you can't solve this problem with only one equation.
 
  • #34
Yep, I used that s=ut+1/2 at^2 for both the horizontal and vertical components ;)
 
  • #35
With 0 acceleration in the horionztal I take it. That's actually what I said in my equations.
 
  • #36
Yes true, but there was still no 1/2 in either of them.

I tried to work it out without the 1/2 and I kept getting a cubic which cannot be solved. The one half made heaps of difference ;)
 
  • #37
OUCH! i simply forgot to write down the half! Pretty amazing that no-one noticed! Sorry about that, dude! :)(the half actually appears from taking the anti-derivative of of t, resulting in 0.5(t)^2.)
 
Last edited:
  • #38
I've haven't learned about anti-derivatives yet. But I see what you have done there.

Its not a problem mate, everyone makes mistakes ;)
 
  • #39
Yup. Those 4 basic equations of motions is what you use when you start out with physics, but you see clearly where they naturally come from first when you learn about derivates and integration.

:)
 

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