A ball starts at 2.9 ft and is thrown up in the air and takes 0.8 seconds to hit the ground what is the highest point it reaches

To find the highest point the ball reaches, we need to use the equations of motion under constant acceleration due to gravity. Let's assume the following:

- The initial height of the ball, \( h_0 = 2.9 \) ft.
- The total time taken for the ball to hit the ground, \( t = 0.8 \) seconds.
- The acceleration due to gravity, \( g = 32 \) ft/s² (downward).

### Step-by-Step Solution:

1. **Determine the initial velocity (\( v_0 \))**:

   The ball takes 0.8 seconds to hit the ground. The total displacement from the initial height to the ground will be \(-2.9\) feet (negative because it ends below the starting point). We can use the equation for displacement:

\[
s = v_0 t - \frac{1}{2} g t^2
\]

Plugging in the values:

\[
-2.9 = v_0 (0.8) - \frac{1}{2} \times 32 \times (0.8)^2
\]

Let's solve for \( v_0 \).

2. **Find the time to reach the highest point**:

At the highest point, the velocity will be zero. We can use the equation:

\[
v = v_0 - gt
\]

Setting \( v = 0 \) (at the highest point):

\[
0 = v_0 - 32 t
\]

Solve for \( t = t_{\text{up}} \), the time to reach the highest point:

\[
t_{\text{up}} = \frac{v_0}{32}
\]

3. **Find the maximum height**:

Use the equation for displacement:

\[
h_{\text{max}} = h_0 + v_

A ball starts at 2.9 ft and is thrown up in the air and takes 0.8 seconds to hit the ground. What is the highest point it reaches?

Find the highest point a ball reaches after being thrown upward from a height of 2.9 ft and taking 0.8 seconds to hit the ground. This detailed solution uses kinematics formulas to calculate the initial velocity, the time to reach the maximum height, and the maximum height itself. Suitable for students in Grades 10-12.

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