Question

A football player kicks the ball with an initial upward velocity of 60 feet per second. This situation is modeled using the equation h=−16t2+60t+0
ℎ
=
−
16
𝑡
2
+
60
𝑡
+
0
, where h is the height in feet and t is the time in seconds. Using this equation, what is the maximum range the ball will reach? Round to the nearest hundredth.

Answer 1

To find the maximum height (not range, since we're dealing with a vertical throw) of the ball, we can use the vertex formula for a quadratic equation in the form \( h(t) = at^2 + bt + c \). In this equation, \( a = -16 \), \( b = 60 \), and \( c = 0 \).

The time at which the maximum height occurs can be found using the formula:

\[
t = -\frac{b}{2a}
\]

Substituting the values of \( a \) and \( b \):

\[
t = -\frac{60}{2 \times -16} = -\frac{60}{-32} = \frac{60}{32} = \frac{15}{8} = 1.875 \text{ seconds}
\]

Next, we will substitute this value of \( t \) back into the height equation \( h(t) \) to find the maximum height:

\[
h = -16(1.875)^2 + 60(1.875) + 0
\]

Calculating \( (1.875)^2 \):

\[
(1.875)^2 = 3.515625
\]

Now substituting that into the equation:

\[
h = -16(3.515625) + 60(1.875)
\]
\[
h = -56.25 + 112.5
\]
\[
h = 56.25
\]

Thus, the maximum height the ball will reach is **56.25 feet**.