Question

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A) Anders throws a baseball straight up from a height of 3 feet above the ground. The initial velocity of
the baseball is 25 feet per second. What is the height of the ball \(1.5\) seconds after it is thrown?
Round your answer to the nearest hundredth.

Answer 1

To determine the height of the baseball \( t \) seconds after being thrown, we can use the equation for the height of an object in free fall. The formula for the height \( h(t) \) above the ground, when an object is thrown straight up, is given by:

\[
h(t) = -16t^2 + v_0t + h_0
\]

where:
- \( v_0 \) is the initial velocity (in feet per second),
- \( h_0 \) is the initial height (in feet),
- \( t \) is the time in seconds,
- The term \(-16t^2\) accounts for the effect of gravity (in feet per second squared).

In your case:
- \( h_0 = 3 \) feet
- \( v_0 = 25 \) feet/second

We want to find \( h(1.5) \):

\[
h(1.5) = -16(1.5)^2 + 25(1.5) + 3
\]

Now we perform the calculations step by step:

1. Calculate \( (1.5)^2 \):
\[
(1.5)^2 = 2.25
\]

2. Calculate \( -16(1.5)^2 \):
\[
-16 \times 2.25 = -36
\]

3. Calculate \( 25(1.5) \):
\[
25 \times 1.5 = 37.5
\]

4. Now substitute these values into the height equation:
\[
h(1.5) = -36 + 37.5 + 3
\]

5. Simplify the equation:
\[
h(1.5) = 1.5 + 3 = 4.5
\]

Thus, the height of the baseball \( 1.5 \) seconds after it is thrown is

\[
\boxed{4.50} \text{ feet}
\]