1. ## SAT height/time problem

$h(t) = c - (d - 4t)^2$

At time $t = 0$, a ball was thrown upward from an initial height of 6 feet. Until the hall hit the ground, its height (in feet), after t seconds was given by function h above, in which $c$ and $d$ are positive constants. If the ball reached its maximum height of 106 ft at $t = 2.5$, what what the height of the ball, in feet, at time $t = 1$?

Obviously, this is a downward sloping parabola since it has a maximum.

So far, I've entered:

$106(2.5) = c - (d - 4(2.5))^2$

but I'm not sure how to proceed from there

2. ## Re: SAT height/time problem

You also know that when t=0 h=6. So 6=c-d^2 . So c=6+d^2. Substitute this for c into your equation and solve for d. Then find c from c=6+d^2.

3. ## Re: SAT height/time problem

Hello, m58!

At time $t = 0$, a ball was thrown upward from an initial height of 6 feet.
Until the hall hit the ground, its height (in feet), after t seconds was given by: / $h(t) \"=\:c - (d-4t)^2$
where $c$ and $d$ are positive constants.

If the ball reached its maximum height of 106 ft at $t = 2.5$,
what is the height of the ball, in feet, at time $t = 1$?

We are told that when $t = 0,\:h = 6$
. . $c - (d - 0)^2 \:=\:6 \quad\Rightarrow\quad c - d^2 \:=\:6\;\;[1]$

We are told that when $t = 2.5,\:h = 106$
. . $c - [d - 4(2.5)]^2 \:=\:106 \quad\Rightarrow\quad c - (d-10)^2 \:=\:106\;\;[2]$

Subtract [1] - [2]: . $-d^2 +(d-10)^2 + d^2 \:=\:6-106 \quad\Rightarrow\quad -20d + 100 \:=\:-100$

. . . . . . . . . . . . . . $-20d \:=\:-200 \quad\Rightarrow\quad d \:=\:10 \quad\Rightarrow\quad c \:=\:106$

Hence: . $h(t) \:=\:106 - (10-4t)^2$

Therefore: . $h(1) \;=\;106 - (10-4)^2 \;=\;106 - 36 \;=\;70\text{ ft}$

4. ## Re: SAT height/time problem

Biffboy, Soroban, you've explained it well! Thank you very much (sorry for the delayed reply!)