Maximum height of a quadratic softball motion

Expert intro: The quadratic function $h(t)=14+32t-16t^2$ reaches its highest point at the vertex.

Detailed walkthrough

Recognize the vertex of the parabola

The height function $h(t)=14+32t-16t^2$ is a quadratic with a negative leading coefficient, so its graph opens downward. That means the maximum height occurs at the vertex, not at an endpoint.

For a quadratic $at^2+bt+c$, the time coordinate of the vertex is

$t=-\frac{b}{2a}.$

Here, $a=-16$ and $b=32$, so

$t=-\frac{32}{2(-16)}=1.$

Evaluate the height at the vertex

Substitute $t=1$ into the function:

$h(1)=14+32(1)-16(1)^2.$

Compute carefully:

$h(1)=14+32-16=30.$

So the softball reaches a maximum height of 30 feet.

Why the answer is 30, not 34

The choice 34 comes from adding the constant 14 and the linear term 32 before accounting for the negative quadratic term. In a motion model like this one, the $-16t^2$ term matters a lot because it represents gravity pulling the ball back down.

You can also confirm the result by noticing the standard projectile form. The parabola rises at first, peaks when $t=1$, and then falls. Since the graph opens downward, the vertex value is the maximum, and that maximum is 30 feet.

💡 Pitfall guide

A frequent mistake with $h(t)=14+32t-16t^2$ is to plug in the wrong time and assume the maximum happens at $t=0$ or at the moment the ball is thrown. That only gives the starting height, not the peak. Another common error is to ignore the negative sign on $-16t^2$ and treat the function like it opens upward, which would turn the vertex into a minimum instead of a maximum. Students also sometimes compute the vertex time as $\frac{b}{2a}$ without the negative sign, which would give $-1$ here and does not make physical sense for a thrown ball. Finally, the answer choice 34 is tempting because it comes from incomplete arithmetic; always substitute the vertex time back into the original function rather than estimating from the coefficients.

🔄 Real-world variant

If the function changed to $h(t)=20+32t-16t^2$, the method would be the same but the maximum height would change. The vertex still occurs at $t=1$ because the coefficients of $t$ and $t^2$ are unchanged. Substituting gives $h(1)=20+32-16=36$. So the modified problem would have a maximum height of 36 feet. This variant is useful because it shows that the vertex time depends on $a$ and $b$, while the maximum height also depends on the starting height term $c$.

🔍 Related terms

vertex of parabola, projectile motion, quadratic function