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