<PRE><B>Question 283805</B>
A water rocket is launched upward with an initial velocity of 48 ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. After how many seconds will the height of the rocket be 20 feet? Show all steps necessary to arrive at your solution.

We need to solve:

20 = 48t - 16t^2

or

16t^2 - 48t + 20 = 0

Divide both sides by 4;

4t^2 - 12t + 5 = 0

(2t - 1)*(2t - 5) = 0

Solutions are:

2t - 1 = 0 and 2t - 5 = 0 or

t = 1/2 and t = 5/2

The first time is on the way up and the second is on the way down.


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