   SEARCH HOME Math Central Quandaries & Queries  Question from kumi, a student: I have made a picture of a football pass 45 degrees with 30yds and height of 7ft and 4ft. But I am just stuck there. I have stared at the worksheet for soo long. I have no idea how to solve this, and I don't even know where this is going (I don't understand how parametric equations work). Can you help me with this problem? 1.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 receiver at a height of 4ft, 30yds directly down field. 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 cosθ)t and y(t) = h + (vo sinθ)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 receiver has to position himself after the quarterback releases the football Hi Kumi.

The parametric equations give you the x and y coordinates of the ball at time t after the ball is released. You haven't given units for time so I am going to assume t is in seconds. So, in particular at time t = 0 seconds (just as the quarterback releases the ball)

x = x(0) = (v0 cosθ) × 0 = 0
y = y(0) = h + (vo sinθ) × 0 - 16 02 = h

Hence at time t = 0 the ball has coordinates (0, h). You know that the ball is released 7 feet of the ground so its y-coordinate is 7 feet. Thus h = 7 feet. θ is the angle from the horizontal at which the ball is released and you are told this is 45 degrees so sinθ = cosθ = 1/√2.

We don't know the time when the receiver catches the ball so I am going to call it t1 seconds. At this time we know that the ball is 4 feet off the ground and 30 yards (90 feet) down field. That is at time t1 the ball has coordinates (90, 4). But we also know that the coordinates are given by the parametric equations so

x(t1) = (vo/√2)t1 = 90
y(t1) = 7 + (vo/√2)t1- 16 t12 = 4

Look closely at the two equations above, (vo/√2)t1 = 90 and (vo/√2)t1 appears in the second equation. Substitute 90 for (vo/√2)t1 in the second equation and solve for t1. Once you have a value for t1 you can substitute it into (vo/√2)t1 = 90 and solve for the initial velocity vo.

I hope this helps,
Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.