[SOLVED] SAT math page 657 #18

**fabxx** · September 28th 2008, 01:34 AM

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

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

Thanks in advance!!

The correct answer is 70 but why?

**Sean12345** · September 28th 2008, 04:21 AM

Hi fabxx,

We have $h(t)=c-(d-4t)^2$ . Note that in the text it states, "at time $t=0$ , a ball was thrown upward from an initial height of $6$ feet" so we can let $t=0$ and $h=6$ . This yields $h(1)=c-d^2=6$ . Call this equation $(*)$ . Also note in the text; " $106$ feet at the time $t=2.5$ " thus letting $t=\frac{5}{2}$ and $h=106$ we have $h\left(\frac{5}{2}\right)=c-\left[d-4\left(\frac{5}{2}\right)\right]^2=c-(d-10)^2=106$ . Define this as equation $(**)$ .

Now we have two equations with two unknowns, namely $c$ and $d$ ;

$c-d^2=6~~~(*)$

$c-(d-10)^2=106~~~(**)$

All that remains is to solve these two equations simultaneously finding $c$ and $d$ then subbing in $t=1$ and the values for $c$ and $d$ . Can you continue?

**fabxx** · September 28th 2008, 04:28 AM

I got d=10 and c=106 and substitute t=1 I got 70 as the final answer!

Thanks Sean12345!

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