Question

a ball is thrown into the air and its position is given by $h(t) = -t^2 + 68t + 20$ where $h$ is the height of the ball in meters $t$ seconds after it has been thrown. find the maximum height reached by the ball and the time at which that happens.

\boxed{} meters

\boxed{} seconds

Explanation:

Step1: Identify the vertex formula for a quadratic function

For a quadratic function in the form \( h(t) = at^2 + bt + c \), the time \( t \) at which the vertex (maximum or minimum) occurs is given by \( t = -\frac{b}{2a} \). In the function \( h(t) = -t^2 + 68t + 20 \), we have \( a = -1 \) and \( b = 68 \).

Step2: Calculate the time at maximum height

Substitute \( a = -1 \) and \( b = 68 \) into the formula for \( t \):

$$ t = -\frac{68}{2(-1)} = -\frac{68}{-2} = 34 $$

So the time at which the maximum height occurs is 34 seconds.

Step3: Calculate the maximum height

Substitute \( t = 34 \) into the height function \( h(t) \):

$$ h(34) = -(34)^2 + 68(34) + 20 $$

First, calculate \( (34)^2 = 1156 \), then \( 68(34) = 2312 \).

$$ h(34) = -1156 + 2312 + 20 = 1156 + 20 = 1176 $$

Answer:

The maximum height is 1176 meters and it occurs at 34 seconds. So the answers are 1176 meters and 34 seconds.