Questions: A ball is launched vertically upward with a velocity of 256 ft / s, so its position function can be modeled as s(t)=-16 t^2+256 t. What is the velocity of the ball when it first reaches a vertical height of 880 ft?

Solution Steps

The position function of the ball is given by:

\[ s(t) = -16t^2 + 256t \]

This function describes the height \( s \) of the ball at any time \( t \).

To find the time when the ball reaches a height of 880 ft, set the position function equal to 880:

\[ -16t^2 + 256t = 880 \]

Rearrange the equation to form a standard quadratic equation:

\[ -16t^2 + 256t - 880 = 0 \]

Divide the entire equation by -16 to simplify:

\[ t^2 - 16t + 55 = 0 \]

Use the quadratic formula to solve for \( t \):

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1 \), \( b = -16 \), and \( c = 55 \).

\[ t = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 55}}{2 \cdot 1} \]

\[ t = \frac{16 \pm \sqrt{256 - 220}}{2} \]

\[ t = \frac{16 \pm \sqrt{36}}{2} \]

\[ t = \frac{16 \pm 6}{2} \]

This gives two possible solutions for \( t \):

\[ t = \frac{22}{2} = 11 \quad \text{and} \quad t = \frac{10}{2} = 5 \]

Since the ball is launched upwards, it first reaches 880 ft at \( t = 5 \) seconds.

The velocity function is the derivative of the position function:

\[ v(t) = \frac{d}{dt}(-16t^2 + 256t) = -32t + 256 \]

Substitute \( t = 5 \) into the velocity function:

\[ v(5) = -32(5) + 256 \]

\[ v(5) = -160 + 256 \]

\[ v(5) = 96 \]

Final Answer