# Question #16ef7

May 3, 2016

Fireworks should explode after $6.25$ seconds.

#### Explanation:

We seek the point (here time $t$) where the quadratic function $s \left(t\right)$ reaches its maximum value. Fortunately we have a simple formula for that:

A quadratic function $s \left(t\right) = a {t}^{2} + b t + c , \setminus \setminus a \ne 0$ reaches its maximum value (if $a < 0$; if $a > 0$ this function has only a global minimum but the following formula holds) at ${t}_{\max} = - \frac{b}{2 a}$. If we're also interested in the value at this point we can simply evaluate it: $s \left({t}_{\max}\right)$.

In our particular fireworks' case we have:
${t}_{\max} = - \frac{200}{2 \cdot \left(- 16\right)} = \frac{25}{4} = 6.25 \setminus \setminus \left[s\right]$

Hence, fireworks should explode after $6.25$ seconds.

If the question was 'What height the fireworks will explode at?' we would also have to evaluate the maximum value: $s \left({t}_{\max}\right) = - 16 \cdot {6.25}^{2} + 200 \cdot 6.25 + 4 = 629 \setminus \setminus \left[m\right]$.

May 3, 2016

$6.25$ seconds.

#### Explanation:

You need to find the height at which the fireworks stop rising and start falling, which you can do by differentiating the equation to find the maximum (the point at which the gradient is zero).

$s ' \left(t\right) = - 32 t + 200$ must equal zero at the highest point.

$32 t = 200$

$t = \frac{200}{32} = 6.25$ secs.