If a ball is thrown into the air at 64 feet per second from the top of a 112​-foot-tall ​building, its height can be modeled by the function Sequals112plus64tminus16tsquared

Question

what is the height three second after it is thrown

Answer 1

To find the height of the ball three seconds after it is thrown, we can use the given function for height:

\[ S(t) = 112 + 64t - 16t^2 \]

where \( S(t) \) represents the height of the ball at time \( t \) in seconds.

We need to evaluate \( S(3) \):

\[ S(3) = 112 + 64(3) - 16(3^2) \]

Calculating each part:

1. Calculate \( 64(3) \): \[ 64 \times 3 = 192 \]

2. Calculate \( 3^2 \): \[ 3^2 = 9 \]

3. Calculate \( 16(9) \): \[ 16 \times 9 = 144 \]

Now substituting these values back into the equation:

\[ S(3) = 112 + 192 - 144 \]

Now perform the addition and subtraction:

1. \( 112 + 192 = 304 \)
2. \( 304 - 144 = 160 \)

Thus, the height of the ball three seconds after it is thrown is:

\[ \boxed{160 \text{ feet}} \]