Question

the function $h(t) = -16t^2 + 100t + 5$ models the height of the ball from the ground at any given second, $t$, and is shown in the
height of the ball (feet)

time (seconds)

estimate when $h(t) = 0$ and explain what it means in the given context.

Explanation:

Step1: Analyze the graph

The graph of \( h(t)= - 16t^{2}+100t + 5\) is a parabola opening downwards. We need to find the values of \( t\) where \( h(t) = 0\), i.e., the x - intercepts of the graph. From the graph, we can see that one of the x - intercepts is near \( t = 0\) (but since time cannot be negative in this context, we consider the positive x - intercept) and the other is near \( t=6.3\) (by visually inspecting the graph, where the parabola crosses the time axis (x - axis) on the positive side).

Step2: Interpret the meaning

In the context of the height of the ball, \( h(t)=0\) means that the height of the ball from the ground is 0 feet. So, when \( h(t) = 0\), the ball is on the ground. The non - negative solution for \( t\) (since time starts from \( t = 0\)) gives the time when the ball hits the ground. From the graph, we estimate that \( t\approx6.3\) seconds (we can also solve the quadratic equation \( - 16t^{2}+100t + 5=0\) using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \( a=- 16\), \( b = 100\), \( c = 5\). \( t=\frac{-100\pm\sqrt{100^{2}-4\times(-16)\times5}}{2\times(-16)}=\frac{-100\pm\sqrt{10000 + 320}}{-32}=\frac{-100\pm\sqrt{10320}}{-32}=\frac{-100\pm101.59}{-32}\). The positive solution is \( t=\frac{-100 - 101.59}{-32}\) (we take the negative sign in the numerator because \(\frac{-100 + 101.59}{-32}\) gives a negative value) \(t=\frac{-201.59}{-32}\approx6.3\) seconds).

Answer:

The ball hits the ground ( \( h(t) = 0\)) at approximately \( t\approx6.3\) seconds. This means the ball is on the ground at about 6.3 seconds after it is launched (or at the start of the motion, the other root near \( t = 0\) might be the time when it is launched from a very low height, but the significant time when it hits the ground is around 6.3 seconds). If we consider the context, the value of \( t\) for which \( h(t)=0\) (and \( t>0\)) is approximately 6.3 seconds, representing the time when the ball returns to the ground.