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.

Re: SAT height/time problem

Hello, m58!

Quote:

At time $\displaystyle t = 0$, 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: /$\displaystyle h(t) \"=\:c - (d-4t)^2$

where $\displaystyle c$ and $\displaystyle d$ are positive constants.

If the ball reached its maximum height of 106 ft at $\displaystyle t = 2.5$,

what is the height of the ball, in feet, at time $\displaystyle t = 1$?

We are told that when $\displaystyle t = 0,\:h = 6$

. . $\displaystyle c - (d - 0)^2 \:=\:6 \quad\Rightarrow\quad c - d^2 \:=\:6\;\;[1]$

We are told that when $\displaystyle t = 2.5,\:h = 106$

. . $\displaystyle c - [d - 4(2.5)]^2 \:=\:106 \quad\Rightarrow\quad c - (d-10)^2 \:=\:106\;\;[2]$

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

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

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

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

Re: SAT height/time problem

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