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 one second after it is thrown

Answer 1

To find the height of the ball one second after it is thrown, we can use the given function for height \( S(t) \):

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

We want to find \( S(1) \), which is the height at \( t = 1 \) second:

\[ S(1) = 112 + 64(1) - 16(1^2) \]

Now, calculating step by step:

1. \( 64(1) = 64 \)
2. \( 16(1^2) = 16 \)

So, substituting these values back into the function:

\[ S(1) = 112 + 64 - 16 \]

Now, adding and subtracting:

\[ S(1) = 112 + 64 - 16 = 112 + 48 = 160 \]

Thus, the height of the ball one second after it is thrown is 160 feet.