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
, 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.(1 point)
feet

Answer 1

To determine the maximum height the football will reach, we can use the vertex formula for a quadratic equation in the standard form \( h = at^2 + bt + c \). In this case, \( a = -16 \), \( b = 60 \), and \( c = 0 \).

The formula to find the time \( t \) at which the maximum height occurs is given by:

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

Plugging in the values of \( a \) and \( b \):

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

Next, we substitute \( t = 1.875 \) back into the height equation \( h = -16t^2 + 60t + 0 \) to find the maximum height:

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

Calculating \( (1.875)^2 \):

\[
(1.875)^2 = 3.515625
\]

Now substitute this value into the height equation:

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

Calculating \( -16(3.515625) \):

\[
-16(3.515625) = -56.25
\]

Calculating \( 60(1.875) \):

\[
60(1.875) = 112.5
\]

Now combine these results:

\[
h = -56.25 + 112.5 = 56.25 \text{ feet}
\]

Thus, the maximum height that the ball will reach is:

\[
\boxed{56.25} \text{ feet}
\]