Applying the Zero Product Property to Solve a Real-World Projectile Problem

We have previously looked at functions representing the motion of projectiles. We analyzed two equivalent expressions that define the same quadratic function, one in quadratic form and the other in factored form.

Now we’re going to look at a similar problem, but with the additional knowledge of how the function relates to the zero product property.

Let’s look at an example.

The equivalent functions below model the height, in feet, of a baseball $t$ seconds after being hit.

$h(t)=-15t^2+44t+3h(t)=(-15t-1)(t-3)$

Using the zero product property, the quadratic expression in factored form allows you to more easily find out when the height of the baseball is 0 feet.

Without graphing, we can determine at what time the baseball reaches the ground.

Let’s apply the zero product property to solve $(-15t-1)(t-3)=0$.

Set each factor equal to 0 and solve.

$\begin{array}{ccccc}\hfill -15t-1& \hfill =\hfill & 0t-3\hfill & \hfill =& \hfill 0\hfill \\ \hfill -15t-1+1& \hfill =\hfill & 0+1t-3+3\hfill & \hfill =& \hfill 0+3\hfill \\ \hfill -15t& \hfill =\hfill & 1t\hfill & \hfill =& \hfill 3\hfill \\ \hfill \frac{-15t}{-15}& \hfill =\hfill & \frac{1}{-15}\hfill \\ \hfill t& \hfill =\hfill & -\frac{1}{15}\hfill \end{array}$

So, $t=-\frac{1}{15}$ and $t=3$. These values of $t$ represent the roots of the function. Because the object was launched at $t=0$, the negative solution doesn’t make sense in this context of time. So the baseball reaches the ground in 3 seconds.

There is one benefit of knowing the function in quadratic form. By substituting 0 in for $t$, we can easily determine at what height the baseball was hit, when $t=0$ seconds.

$h(t)=-15t^2+44t+3$

$h(0)=-15(0{)}^2+44(0)+3$

$h(0)=3$

So, the baseball was hit at a height of 3 feet. Note that this point, $(0,3)$, is the $y$-intercept of the function when graphed.