The equation $h(t) = -16t^2 + 48t + 40$ is a quadratic. For a downward-opening parabola, where on the graph does the maximum occur?

For a quadratic $y = at^2 + bt + c$, the $t$-coordinate of the vertex is given by $t = -\frac{b}{2a}$. Identify $a$ and $b$ from the given equation and then compute this value.

Make sure you include both the negative sign and the $2$ in the formula $t = -\frac{b}{2a}$, and simplify the fraction step by step.

In Desmos, type h(t) = -16t^2 + 48t + 40 (you can also use y and x instead of h and t: y = -16x^2 + 48x + 40).

Adjust the window if needed so you can clearly see the top part of the parabola (where it curves downward and reaches a highest point).

Click on the graph near its highest point; Desmos will mark the vertex. Note the $t$- (or $x$-) coordinate of this maximum point — that value is the time when the ball reaches its maximum height.

The height is given by a quadratic function

Because the coefficient of $t^2$ is negative ($-16$), the parabola opens downward, so it has a maximum value at its vertex (highest point). The question is asking for the time coordinate of this vertex.

For a quadratic written as

the $t$-coordinate of the vertex (and thus the time when the function reaches its maximum for a downward-opening parabola) is

Here, $a=-16$ and $b=48$.

Substitute $a=-16$ and $b=48$ into $t = -\frac{b}{2a}$:

Now divide $48$ by $32$:

So, according to the model, the ball reaches its maximum height $1.5$ seconds after launch.