An object is launched vertically in the air at 36.75 meters per second from a 11​-meter-tall platform. Using the projectile motion model h left parenthesis t right parenthesis equals negative 4.9 t squared plus v 0 t plus h 0 comma where​ h(t) is the height of the projectile t seconds after its​ departure, v 0 is the initial velocity in meters per​ second, and h 0 is the initial height in​ meters, determine how long it will take for the object to reach its maximum height. What is the maximum​ height?
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Part 1
The object will reach its maximum height in

enter your response here seconds.

To determine how long it will take for the object to reach its maximum height and what that maximum height will be, we will use the given projectile motion model:
\[ h(t) = -4.9t^2 + v_0 t + h_0 \]

Here, \( v_0 = 36.75 \, \text{m/s} \) (initial velocity) and \( h_0 = 11 \, \text{m} \) (initial height).

### Part 1: Time to Reach Maximum Height

The time to reach maximum height can be found by determining when the velocity is zero (i.e., the object stops ascending). The velocity function is the derivative of the height function \( h(t) \):
\[ v(t) = \frac{dh(t)}{dt} = -9.8t + v_0 \]

Set the velocity equal to zero to find the time \( t \) at which the object reaches its maximum height:
\[ 0 = -9.8t + 36.75 \]
\[ 9.8t = 36.75 \]
\[ t = \frac{36.75}{9.8} \]
\[ t \approx 3.75 \, \text{seconds} \]

Therefore, the object will reach its maximum height in approximately \( 3.75 \) seconds.

### Part 2: Maximum Height

To find the maximum height, substitute \( t = 3.75 \) seconds back into the height function \( h(t) \):
\[ h(3.75) = -4.9(3.75)^2 + 36.75(3.75) + 11 \]

Calculating each term:
\[ -4.9(3.75)^2 = -4.9 \times 14.0625 \approx -68.90625 \]
\[ 36.75(3.75) = 137.8125 \]
\[ 11 = 11 \]

Summing these up:
\[ h(3.75) = -68.90625 + 137.8125 + 11 \]
\[ h(3.75) \approx 79.90625 \, \text{meters} \]

Therefore, the maximum height of the object is approximately \( 79.91 \) meters.

### Summary
- The object will reach its maximum height in approximately \( 3.75 \) seconds.
- The maximum height of the object is approximately \( 79.91 \) meters.

Would you like more details or have any questions about this solution?

Here are some related questions you might find interesting:
1. How do you find the time of flight for the object?
2. What will be the object's height at \( t = 2 \) seconds?
3. How do you determine the velocity of the object at any given time \( t \)?
4. What is the total time the object spends in the air before hitting the ground?
5. How do you find the speed of the object just before it hits the ground?
6. What is the significance of the coefficient -4.9 in the height function?
7. How would the problem change if the initial height was different?
8. How do you derive the equations for projectile motion from basic physics principles?

**Tip:** When solving projectile motion problems, always start by identifying and writing down the known values and the equations that apply to the scenario. This helps organize your solution process and ensures accuracy.

Explore how to calculate the maximum height and time to reach it for an object launched vertically at 36.75 m/s from an 11-meter-tall platform using projectile motion equations. Suitable for high school students studying kinematics and physics.

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