



 
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)
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 ycoordinate 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 t_{1} 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 t_{1} the ball has coordinates (90, 4). But we also know that the coordinates are given by the parametric equations so
Look closely at the two equations above, (vo/√2)t_{1} = 90 and (vo/√2)t_{1} appears in the second equation. Substitute 90 for (vo/√2)t_{1} in the second equation and solve for t_{1}. Once you have a value for t_{1} you can substitute it into (vo/√2)t_{1} = 90 and solve for the initial velocity vo. I hope this helps,  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 