A projectile is launched upward with a velocity of 288
 feet per second from the top of a 35
-foot platform.  What is the maximum height attained by the projectile?



To find the maximum height attained by the projectile, we can use the formula for the height of a projectile:

\[
h(t) = -16t^2 + v_0 t + h_0
\]

where:
- \( h(t) \) is the height of the projectile at time \( t \),
- \( v_0 \) is the initial velocity (288 feet per second),
- \( h_0 \) is the initial height (35 feet),
- \( -16 \) is the acceleration due to gravity in feet per second squared.

### Step 1: Write the equation for the height
Given the values:
\[
h(t) = -16t^2 + 288t + 35
\]

### Step 2: Determine the time at which the maximum height occurs
The time at which the maximum height occurs can be found by using the formula for the vertex of a parabola, which is given by:
\[
t_{\text{max}} = \frac{-b}{2a}
\]
where \( a = -16 \) and \( b = 288 \).

\[
t_{\text{max}} = \frac{-288}{2(-16)} = \frac{-288}{-32} = 9 \text{ seconds}
\]

### Step 3: Calculate the maximum height
Now, substitute \( t = 9 \) seconds back into the height equation:

\[
h(9) = -16(9)^2 + 288(9) + 35
\]

\[
h(9) = -16(81) + 2592 + 35
\]

\[
h(9) = -1296 + 2592 + 35 = 1331 \text{ feet}
\]

Thus, the maximum height attained by the projectile is **1331 feet**.

Would you like more details or have any questions?

### Relative Questions:
1. How would the maximum height change if the initial velocity was different?
2. What is the significance of the coefficient \(-16\) in the height equation?
3. How does the initial height (\(h_0\)) affect the maximum height?
4. Can you explain how the time to reach the maximum height is calculated?
5. What would happen to the maximum height if the projectile was launched from a higher platform?

**Tip:** Always remember that the time to reach the maximum height in projectile motion can be found using the vertex formula, which comes from the symmetry of the parabolic path.

Learn how to calculate the maximum height attained by a projectile launched upward with an initial velocity of 288 feet per second from a 35-foot platform. Detailed step-by-step solution using projectile motion and quadratic equations.

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