My teacher gave us a brainteaser for extra points... haha but it's been a long time since i used parabolas... guide me to the solution please!! Thank You guys!
Given equation: h(t)=c-(d-4t)^2
At time t=0 a ball was thrown upward from an initial height og 6 feet. Until the ball hit the ground, tis height, in feet, after t secons was given by the function h above, in which d and c are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the heigh in feet of the ball at time t=1?
Well, I supposed that c=6.. .but that won't make sense when you plug in 2.5 to t.( 6-d^2-100) I am seriously lost.....
We know two points on the parabola in the form (t,h(t)): (0,6), the initial point and (2.5,106), the vertex (maximum)
Putting the parabola in the form h(t) = a(t-h)^2+k using the point (0,6) for (t,h(t)) and the point (2.5,106) for (h,k) yields:
solve for a:
Now we know our standard equation is h(t)=-16(t-2.5)^2+106
Replace t with 1 to find the height at 1 second:
The height at 1 second is 70 feet.
I'm not sure why we would need to know the first equation that was given to you. Just using the vertex and initial height at t=0 would suffice to find the height at t=1 second.