Projectile Motion - Finding Ini. Vel. using Range and Angle

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Homework Help Overview

The problem involves projectile motion, specifically calculating the initial velocity required for a quarterback to throw a football to a stationary receiver at a distance of 31.5 meters, given an angle of 40 degrees. The context includes the use of the range equation and considerations regarding the sign of gravitational acceleration.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the range equation and the implications of gravity being negative. Some explore the derivation of the equation and the relationship between horizontal and vertical motion.

Discussion Status

Participants are actively engaging with the problem, clarifying the use of the range equation and addressing concerns about the sign of gravity. There is an exchange of ideas regarding the derivation of equations and the reasoning behind their application.

Contextual Notes

There is a mention of the need to derive equations rather than relying solely on memorization, indicating a focus on understanding the underlying principles of projectile motion.

Tearsandrille
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Homework Statement


A quarterback throws the football to a stationary receiver who is 31.5m down the field. If the football is thrown at an initial angle of 40.0 degrees to the ground, at what initial speed must the quarterback throw the ball for it to reach the receiver, if the ball is caught at the same level it was thrown?


Homework Equations


R = (Vo2*sin(2*theta))/g


The Attempt at a Solution


The issue I am having is that if gravity is negative I don't get a real number, but I can not understand or explain why gravity would be positive.
 
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Hi Tearsandrille! :smile:

(have a theta: θ :wink:)
Tearsandrille said:
R = (Vo2*sin(2*theta))/g

Did you just copy that equation from a book? :redface:

Yes it's correct, but you'll never be able to remember equations like that in the exam, and anyway you need to be able to prove them yourself …

solve the simultaneous equations for the x and y directions (eliminating t), and you'll see why it doesn't matter that g is negative. :wink:
 
No, I did not just copy the equation from the book. I did derive it using the fact that R = Vxinitial*t and t = (2Vosin θ)/g (which I derived from t = (Vf - Vi)/-g)

So, the reason that I don't need to worry about the negative when I use the range equation is that I already compensated for that when I solved for t in the equation Vf=Vi + gt.

Thank you.
 
Tearsandrille said:
So, the reason that I don't need to worry about the negative when I use the range equation is that I already compensated for that when I solved for t in the equation Vf=Vi + gt.

Yup! :smile:
 

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