Eliminating the parameter, physics equation.

In summary, the question asks for the speed of a football that is released by a quarterback at a height of 7 feet and caught by a receiver at a height of 4 feet, 30 yards downfield. The pass is released at an angle of 35 degrees with the horizontal. Using the given parameters of x = (vi*cosθ)t and y = h + (vi*sinθ)t - 16t^2, the solution involves solving for t and vi using the known values for h, y, and x.
  • #1
Xorlev
4
0
[SOLVED] Eliminating the parameter, physics equation.

The question reads "The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of 35 degrees with the horizontal. Find the speed of the football when it's released.

I'm given x = (vi*cosθ)t and y = h + (vi*sinθ)t - 16t^2 as my parameters.

I need a pointer in the right direction. I attempted to equate it to cos 35 and sin 35, but I wasn't sure how to solve vi from there. Thank you for your help.
 
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  • #2
x = (vi*cosθ)t
t=x/(vi*cosθ)

try subbing that into y and you'll get y= function in terms of x
 
  • #3
Since it's in the precalc board I'll assume you don't know basic motion equations, so...

Look at your unknowns:
x displacement (change in x)
initial velocity
time
initial height
final height (y)

Reading the problem can you figure out what you know versus what you are looking for?
 
Last edited:
  • #4
Xorlev said:
The question reads "The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of 35 degrees with the horizontal. Find the speed of the football when it's released.

I'm given x = (vi*cosθ)t and y = h + (vi*sinθ)t - 16t^2 as my parameters.

I need a pointer in the right direction. I attempted to equate it to cos 35 and sin 35, but I wasn't sure how to solve vi from there. Thank you for your help.
What do you mean "equate it to cos 35 and sin 35"? Equate what? Obviously [itex]\theta= 35[/itex] since that is the only angle you are given so, if you meant replace [itex]cos(\theta])[/itex] by cos(35) and [itex]sin(\theta)[/itex] by sin(35), yes, that is what you do. You also know that h= 7 (feet above the playing field), y= 4 feet, x= 30 yards= 90 feet. If you put those numbers into your two equations you will have two equations with the two unknown values vi and t. You should be able to solve two equations for two unknowns. vi is, of course, the value you want.
 
  • #5
I ended up solving for t by using vi*t = (90/cos(35)), substituting it into y = h + (vi*sinθ)t - 16t^2, and solving for t, then solving for vi with vi = (90/(cos(35)*t). Thank you for the help, though.
 

1. What is the purpose of eliminating the parameter in physics equations?

The purpose of eliminating the parameter in physics equations is to simplify the equation and make it easier to solve. It allows us to express the relationship between different variables in a more concise and understandable form.

2. How do you eliminate the parameter in a physics equation?

The process of eliminating the parameter in a physics equation involves isolating the parameter on one side of the equation and expressing it in terms of the other variables. This can be done by using algebraic manipulations and basic mathematical operations.

3. Why is eliminating the parameter important in physics?

Eliminating the parameter in physics equations is important because it helps us to understand the relationship between different variables more clearly. It also allows us to make predictions and solve problems using the equation without having to consider the effect of the parameter.

4. Can the parameter be eliminated in all physics equations?

No, not all physics equations can have their parameters eliminated. It depends on the form and complexity of the equation. Some equations may require the parameter to be present in order to accurately represent the relationship between the variables.

5. Are there any drawbacks to eliminating the parameter in physics equations?

One potential drawback of eliminating the parameter in physics equations is that it may result in a loss of information. The parameter may contain important physical meaning and removing it could limit our understanding of the system being studied. Additionally, the process of eliminating the parameter may lead to a less accurate solution if done incorrectly.

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