Solving Quadratic Equations

Quote:

Example 3: When a football is kicked with a vertical speed of 20m/s, its height, *h* metres, after *t* seconds is given by the formula:

$\displaystyle h=20t-5t^2$

How long after the kick is the football at a height of 15m?

Solution

Substitute 15 for *h* in the formula:

$\displaystyle 15=20t-5t^2$

This produces a quadratic equation. The solution of the equation gives the times when the height is 15m.

Collect all the terms on one side of the equation.

$\displaystyle 5t^2-20t+15=0$

$\displaystyle 5(t^2-4t+3)=0$

$\displaystyle t^2-4t+3=0$

$\displaystyle (t-1)(t-3)=0$

Either

$\displaystyle t-1 = 0$

$\displaystyle t=1$

or

$\displaystyle t-3=0$

$\displaystyle t=3$

The football is at a height of 15m twice: first on the way up, 1 s after the kick, then on the way down, 3 s after the kick.

So I got this question:

A baseball is hit with a vertical speed of 31.3m/s. Use the formula in the modelling box following Example 3. How long after being hit is the ball at a height of 25m? Give your answer to 1 decimal place.

How exactly do I form the equation? I got it but I don't think I'm on the right track. Thanks a bunch.