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=−5x2+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 equation \( y = -5x^2 + 11x + 7 \).

Let's calculate it step by step:

1. Substitute \( x = 2 \) into the equation:
\[
y = -5(2)^2 + 11(2) + 7
\]

2. Calculate \( (2)^2 \):
\[
(2)^2 = 4
\]

3. Substitute \( 4 \) into the equation:
\[
y = -5(4) + 11(2) + 7
\]

4. Calculate \( -5(4) \):
\[
-5(4) = -20
\]

5. Calculate \( 11(2) \):
\[
11(2) = 22
\]

6. Now substitute these values back into the equation:
\[
y = -20 + 22 + 7
\]

7. Combine the constants:
\[
y = -20 + 22 = 2
\]
\[
y = 2 + 7 = 9
\]

Thus, the height of the ball after 2 seconds is \( \boxed{9} \) meters.