Answer:
Step-by-step explanation:
The pull of gravity in feet per second squared is 32. Going backwards from that (which requires integration in calculus) gives us eventually the position function which has a leading coefficient of -16t-squared. The v0 is our initial vertical velocity. The position equation for this situation then is
[tex]s(t)=-16t^2+112t[/tex]
We are asked to find at what times (yes, there are 2 of them!) when the position of the projectile will be at 192 feet. Because of the nature of a parabola, the object will pass a certain number of feet going up, but then because of gravity, will pass by that same number of feet coming back down. Our position of 192 feet is subbed in for s(t) to give us
[tex]192=-16t^2+112t[/tex] and we are asked to solve for t. The only way we can do that is to put everything on one side of the equals sign and set the quadratic equal to 0 and factor. At the same time I am going to factor out a -16 to make the numbers a bit more manageable when we use the quadratic formula.
[tex]-16(t^2-7t+12)=0[/tex]
By the Zero Product Property, either -16 = 0 which of course it doesn't!, or
[tex]t^2-7t+12=0[/tex]
Factor that using the quadratic formula or whatever method you've been taught that is the best for you and get the times that the object is 192 feet in the air is at 3 seconds and again at 4 seconds.
Along the lines of parabolic motion, if the object passes 192 at 3 and then again at 4 seconds, that means that somewhere in between 3 and 4 seconds it reached its max height. To be exact, at 3.5 seconds. To find the max height of the object, sub 3.5 in for t in the original position equation to get the height at 3.5 seconds:
[tex]s(3.5)=-16(3.5)^2+112(3.5)[/tex] to get that the max height of the projectile is 196 feet. That's just a bit extra. I am a teacher after all of both calculus and physics and can't let a teaching moment go by!! ;)
