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)

Answer 1

To find the maximum height the football reaches, we can use the equation given:

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

This is a quadratic equation of the form \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 60 \), and \( c = 0 \).

The maximum height of a parabola (which opens downward in this case) occurs at the vertex. The time \( t \) at which the maximum height occurs can be found using the formula:

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

Substituting in our values for \( a \) and \( b \):

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

Now, we can plug this value of \( t \) back into the height equation to find the maximum height \( h \):

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

Calculating \( (1.875)^2 \):

\[ (1.875)^2 = 3.515625 \]

Now substitute it back into the height equation:

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

Calculating each term:

1. \(-16(3.515625) = -56.25\)
2. \(60(1.875) = 112.5\)

Now combine these results:

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

Thus, the maximum height the ball will reach is:

\[ \boxed{56.25} \] feet.