Expert: Paul Klarreich - 3/5/2008

Question
The quarterback of a football team releases a pass at a height, h, of 7ft above the playing field, and the football is caught by a reciever at a height of 4ft, 30yds directly downfield. The pass is released at an angle of 45 degrees with the horizontal with an initial velocity of vo. The parametric equations for the position of the football at time t are given, in general, by x(t)=(vo cosč)t and y=h+(vo sinč)t-16t^2.

a. find the initial velocity of the football when it is released
b. write the specific set of parametric equations for the path of the football
c.use a graphing calculator to graph the path of the football and approximate its maximum height
d. find the time the reciever has to position himself after the quarterback releases the football

Answer
Questioner:   Kumi
Category:  Advanced Math
Private:  No
 
Subject:  Parametric equations
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Hi, Kumi,

Your Question:  Eli Manning releases a pass at a height, h, of 7ft above the playing field, and the football is caught by a reciever

>> that's receiver.

at a height of 4ft, 30yds directly downfield.

>> that's 120 feet -- let's keep units consistent.

The pass is released at an angle of 45 degrees with the horizontal

>> smart -- that always gives maximum distance.  [That's another calculus question.]

with an initial velocity of vo. The parametric equations for the position of the football at time t are given, in general, by x(t)=(vo cos 45)t and y=h+(vo sin 45)t-16t^2.

>> I put in the 45 degrees there.  Don't use special symbols on this site -- they get eaten alive.

a. find the initial velocity of the football when it is released

b. write the specific set of parametric equations for the path of the football

c.use a graphing calculator to graph the path of the football and approximate its maximum height

>> I'll see what I can do.

d. find the time the recEIver has to position himself after the quarterback releases the football


a. That would be v0.  And you have  h = 7.  So your equations are:

x(t)=(vo cos 45)t and y = 7 + (vo sin 45)t - 16t^2.

And, putting in:

x(t)=(vo sqrt(2)/2)t and y=h+(vo sqrt(2)/2)t-16t^2.

So, at some time t, when the recEIver catches it, you have:

y = 4 ft. [That's not Plaxico Burress, is it?  Okay, he catches it at knee level -- that's about 4 ft.]

x = 120 ft.

Put those in:

120 = (vo sqrt(2)/2)t

4 = 7 + (vo sqrt(2)/2)t - 16t^2.

So you have two simultaneous equations.

4 = 7 + 120 - 16t^2.

0 = 123 - 16t^2.

t = sqrt(123)/4

The rest should be fairly easy.  That 't' is your answer to (d), and you can now use:

120 = (vo sqrt(2)/2)t

120 = (vo sqrt(2)/2)sqrt(123)/4

to solve for  v0 and write your equations.  Use your (non-graphing) calculator to approximate t and v0.