Question

A ball is thrown straight up from 7 meters above the ground. The height y (in meters) of the ball after x seconds is modeled by the equation

y=-5x^2+11x+7

What is the height of the ball after 2 seconds?

Answer 1

To find the height of the ball after 2 seconds, we can substitute \( x = 2 \) into the given equation:

\[
y = -5x^2 + 11x + 7
\]

Substituting \( x = 2 \):

\[
y = -5(2)^2 + 11(2) + 7
\]

Now, calculate \( (2)^2 \):

\[
(2)^2 = 4
\]

So we have:

\[
y = -5(4) + 11(2) + 7
\]

Calculating each term:

\[
-5(4) = -20
\]
\[
11(2) = 22
\]

Now we put it all together:

\[
y = -20 + 22 + 7
\]

Calculating:

\[
-20 + 22 = 2
\]
\[
2 + 7 = 9
\]

Thus, the height of the ball after 2 seconds is:

\[
\boxed{9} \text{ meters}
\]