Questions: A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=128+112t-16t^2. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down?

Solution Steps

To find when the ball strikes the ground, we need to determine when the distance \( s \) from the ground is zero. The equation for the distance is given by:

\[ s = 128 + 112t - 16t^2 \]

Set \( s = 0 \) to find the time \( t \) when the ball hits the ground:

\[ 0 = 128 + 112t - 16t^2 \]

Rearrange the equation to standard quadratic form:

\[ 16t^2 - 112t - 128 = 0 \]

Divide the entire equation by 16 to simplify:

\[ t^2 - 7t - 8 = 0 \]

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -7 \), and \( c = -8 \):

\[ t = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]

\[ t = \frac{7 \pm \sqrt{49 + 32}}{2} \]

\[ t = \frac{7 \pm \sqrt{81}}{2} \]

\[ t = \frac{7 \pm 9}{2} \]

This gives two solutions:

\[ t = \frac{16}{2} = 8 \quad \text{and} \quad t = \frac{-2}{2} = -1 \]

Since time cannot be negative, the ball strikes the ground at \( t = 8 \) seconds.

The ball passes the top of the building when \( s = 128 \) on its way down. We need to find the second time \( t \) when \( s = 128 \):

\[ 128 = 128 + 112t - 16t^2 \]

Simplify the equation:

\[ 0 = 112t - 16t^2 \]

Factor out \( t \):

\[ t(112 - 16t) = 0 \]

This gives two solutions:

\[ t = 0 \quad \text{or} \quad 112 - 16t = 0 \]

Solve for \( t \):

\[ 16t = 112 \]

\[ t = \frac{112}{16} = 7 \]

The ball passes the top of the building again at \( t = 7 \) seconds.

Final Answer