# Eliminating the parameter, physics equation.

1. Nov 19, 2007

### Xorlev

[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.

2. Nov 19, 2007

### rock.freak667

x = (vi*cosθ)t
t=x/(vi*cosθ)

try subbing that into y and you'll get y= function in terms of x

3. Nov 19, 2007

### Feldoh

Since it's in the precalc board I'll assume you don't know basic motion equations, so...

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: Nov 19, 2007
4. Nov 20, 2007

### HallsofIvy

Staff Emeritus
What do you mean "equate it to cos 35 and sin 35"? Equate what? Obviously $\theta= 35$ since that is the only angle you are given so, if you meant replace $cos(\theta])$ by cos(35) and $sin(\theta)$ 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. Nov 20, 2007

### Xorlev

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.