# Math Help - equation of ball

1. ## equation of ball

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

Problem:
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.....

2. Originally Posted by jjkins
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

Problem:
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.....
I think the key to solve this question is to find the positive constants c and d. Let’s consider the given height function $h(t)= c-(d-4t)^2$, realize that $(d-4t)^2\ge 0$, so the maximum height can only be obtained when $(d-4t)^2= 0$, which means when $t=\frac{d}{4}$ it reaches the maximum height (and the maximum height is $h_{\mathtt{max}}=c$). Now you found $c=106$ and $\frac{d}{4}=2.5\Rightarrow d=10$, then you can answer the question by finding $h(1)=106-(10-4)^2$

Roy

3. Originally Posted by jjkins
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

Problem:
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.....
h(t) = c-(d-4t)^2

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:

6=a(0-2.5)^2+106

solve for a:

a=-16

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:

h(1)=-16(1-2.5)^2+106
h(1)=70

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.