Question

a professional baseball player throws a baseball from the outfield nearly 400 feet to home plate. the height of the thrown baseball above the ground, in feet, is modeled by the function $h(t) = -16t^2 - 62.25t + 7$, where $t$ is the number of seconds since the ball left the players hand. a graph of the function is shown.

in this context, what is the domain of the function?

Explanation:

Step1: Understand Domain in Context

The domain of a function in a real - world context (here, the time \(t\) when the baseball is in flight) must be non - negative (since time starts at \(t = 0\) when the ball is thrown) and end when the ball hits the ground (height \(h(t)=0\)).

Step2: Solve \(h(t)=0\) for \(t\)

We have the quadratic function \(h(t)=- 16t^{2}-62.25t + 7\). The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a=-16\), \(b=-62.25\), and \(c = 7\).

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-62.25)^{2}-4\times(-16)\times7\)
\(=3875.0625 + 448=4323.0625\)

Then, \(t=\frac{-(-62.25)\pm\sqrt{4323.0625}}{2\times(-16)}=\frac{62.25\pm65.75}{-32}\)

We have two solutions:
- \(t_1=\frac{62.25 + 65.75}{-32}=\frac{128}{-32}=- 4\) (discarded since time can't be negative)
- \(t_2=\frac{62.25-65.75}{-32}=\frac{-3.5}{-32}\approx0.109\)? Wait, this seems wrong. Wait, maybe I made a mistake in the sign of \(b\). Wait, the function is \(h(t)=-16t^{2}-62.25t + 7\), so when \(h(t) = 0\), \(16t^{2}+62.25t - 7=0\) (multiplying both sides by - 1). Now \(a = 16\), \(b = 62.25\), \(c=-7\)

Discriminant \(\Delta=b^{2}-4ac=(62.25)^{2}-4\times16\times(-7)=3875.0625 + 448 = 4323.0625\)

\(t=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-62.25\pm\sqrt{4323.0625}}{32}\)

\(\sqrt{4323.0625}=65.75\)

So \(t=\frac{-62.25 + 65.75}{32}=\frac{3.5}{32}\approx0.109\) (discarded, since when \(t = 0\), \(h(0)=7\), and the graph shows that the ball goes up and then comes down. Wait, maybe the original function has a sign error? Wait, the height function of a projectile is usually \(h(t)=-16t^{2}+v_0t + h_0\), where \(v_0\) is the initial velocity. In the given function \(h(t)=-16t^{2}-62.25t + 7\), the coefficient of \(t\) is negative, which would mean the initial velocity is negative (the ball is thrown downward), which doesn't make sense. But maybe in the context of the graph, we can look at the \(t\) - intercepts. From the graph, the ball is thrown at \(t = 0\) (since at \(t = 0\), \(h(0)=7\)) and lands at some positive \(t\). Wait, maybe I misread the function. Wait, the problem says "the height of the thrown baseball above the ground", so \(t\) starts at \(t = 0\) (when the ball is thrown) and ends when the ball hits the ground (when \(h(t)=0\)). But from the graph, we can see that the left - hand side of the parabola (for negative \(t\)) is not in the context, and the right - hand \(t\) - intercept is around, looking at the graph, the \(t\) - axis is marked with - 10, - 5, 0, 5, 10. Wait, maybe the function was written incorrectly, but in the context, the domain is from \(t = 0\) (when the ball is thrown) to the time when the ball hits the ground (the positive \(t\) - intercept). But let's re - evaluate.

Wait, maybe the function is \(h(t)=-16t^{2}+62.25t + 7\) (a typo with the sign of the linear term). Let's assume that. Then \(a=-16\), \(b = 62.25\), \(c = 7\)

Discriminant \(\Delta=(62.25)^{2}-4\times(-16)\times7=3875.0625 + 448=4323.0625\)

\(t=\frac{-62.25\pm\sqrt{4323.0625}}{2\times(-16)}=\frac{-62.25\pm65.75}{-32}\)

First solution: \(t=\frac{-62.25 + 65.75}{-32}=\frac{3.5}{-32}\approx - 0.109\) (discarded)

Second solution: \(t=\frac{-62.25-65.75}{-32}=\frac{-128}{-32}=4\)

Ah, that makes sense. So probably there was a sign error in the original function, and the correct function should have a positive linear term. So if we take \(t\) from \(0\) to \(4\) (since at \(t = 4\), \(h(4)=-16\times16+62.25\times4 + 7=-256 + 249+7 = 0\)). And since time starts at \(t = 0\) (when the ball is thrown) and ends at \(t = 4\) (when the ball hits the ground), the domain is \(0\leq t\leq4\) (or \(0\leq t<4\) or \(0\leq t\leq4\) depending on if we consider the exact moment it hits the ground).

Looking at the graph, the \(t\) - axis has marks at - 10, - 5, 0, 5, 10. The parabola crosses the \(t\) - axis at \(t = 0\) (wait, no, at \(t = 0\), \(h(0)=7\)) and at some positive \(t\) around \(t = 4\) (if we consider the corrected function). So the domain in context is the set of all real numbers \(t\) such that \(t\) is between \(0\) and the time when the ball hits the ground, so \(0\leq t\leq4\) (or approximately \(0\leq t\leq4\)).

Answer:

The domain of the function in this context is \(0\leq t\leq4\) (or the interval \([0, 4]\) where \(t\) is in seconds).