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


=

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.(1 point)
feet

1 answer

To find the maximum height (or range) the ball will reach using the equation \( h = -16t^2 + 60t + 0 \), we need to determine the vertex of the parabola represented by this quadratic equation. The maximum height occurs at the vertex, which can be found using the formula for the time \( t \) at which the vertex occurs:

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

where \( a \) and \( b \) are coefficients from the quadratic equation \( h = at^2 + bt + c \).

In this case, the coefficients are:

  • \( a = -16 \)
  • \( b = 60 \)
  • \( c = 0 \)

Substituting \( a \) and \( b \) into the vertex formula:

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

Next, we need to find the maximum height by substituting this value of \( t \) back into the height equation:

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

Calculating \( (1.875)^2 \):

\[ (1.875)^2 = 3.515625 \]

Now substitute this value back into the height equation:

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

Calculating each term:

\[ h = -56.25 + 112.5 \]

Now combine the terms:

\[ h = -56.25 + 112.5 = 56.25 \]

Thus, the maximum height (or range) the ball will reach is

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

Rounding to the nearest hundredth, the answer remains \( 56.25 \) feet.