A bullet is fired upward from ground level. Its height above the ground​ (in feet) at time t seconds is given by the following equation.
​H(t)equalsnegative 16 t squared plus 736 t
​(a) Find the vertex.
​(b) Find the time at which the bullet reaches its maximum height.
​(c) Find the maximum height.
​(d) Find the time at which the bullet hits the ground​ (when the height is equal to​ zero).

We are given the height equation for the bullet:  
\[ H(t) = -16t^2 + 736t \]  
where \( H(t) \) represents the height (in feet) at time \( t \) seconds.

### (a) Find the vertex:
The equation \( H(t) = -16t^2 + 736t \) is in the form of a quadratic equation:  
\[ H(t) = at^2 + bt + c \]
where \( a = -16 \), \( b = 736 \), and \( c = 0 \).

The vertex of a parabola represented by a quadratic equation \( y = ax^2 + bx + c \) occurs at \( t = \frac{-b}{2a} \).

Using this formula:
\[
t = \frac{-736}{2(-16)} = \frac{-736}{-32} = 23 \, \text{seconds}
\]

Thus, the vertex occurs at \( t = 23 \) seconds.

### (b) Find the time at which the bullet reaches its maximum height:
Since the parabola opens downward (because \( a = -16 \) is negative), the vertex represents the maximum point. From part (a), we already found that the time of the maximum height is \( t = 23 \) seconds.

### (c) Find the maximum height:
To find the maximum height, we substitute \( t = 23 \) into the height equation:
\[
H(23) = -16(23)^2 + 736(23)
\]
First, calculate \( 23^2 = 529 \):
\[
H(23) = -16(529) + 736(23) = -8464 + 16928 = 8464 \, \text{feet}
\]

Thus, the maximum height is 8464 feet.

### (d) Find the time at which the bullet hits the ground:
The bullet hits the ground when \( H(t) = 0 \), so we set the height equation equal to zero and solve for \( t \):
\[
-16t^2 + 736t = 0
\]
Factor out \( t \):
\[
t(-16t + 736) = 0
\]
This gives two solutions:
\[
t = 0 \quad \text{or} \quad -16t + 736 = 0
\]
Solving \( -16t + 736 = 0 \):
\[
-16t = -736 \quad \Rightarrow \quad t = \frac{736}{16} = 46
\]

So, the bullet hits the ground at \( t = 46 \) seconds.

### Summary of answers:
- **(a)** The vertex is at \( t = 23 \).
- **(b)** The bullet reaches its maximum height at \( t = 23 \) seconds.
- **(c)** The maximum height is 8464 feet.
- **(d)** The bullet hits the ground at \( t = 46 \) seconds.

Would you like further clarification or details on any part of this?  

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Here are 5 related questions for further practice:
1. How would the equation change if the initial velocity was 800 feet per second instead of 736 feet per second?
2. What would happen to the maximum height if gravity was \( -32 \, \text{ft/s}^2 \) instead of \( -16 \, \text{ft/s}^2 \)?
3. Can you solve for the bullet's height after 10 seconds?
4. How would you find the total flight duration for a bullet fired at an angle with the same initial speed?
5. What is the average velocity of the bullet over its entire flight?

**Tip:** In projectile motion, the time to reach the maximum height is always half the total flight time.

A bullet is fired upward from ground level. Its height above the ground (in feet) at time t seconds is given by the following equation: H(t) = -16t^2 + 736t. (a) Find the vertex. (b) Find the time at which the bullet reaches its maximum height. (c) Find the maximum height. (d) Find the time at which the bullet hits the ground (when the height is equal to zero).

Explore how to solve a projectile motion problem where a bullet is fired upward from the ground. Learn to calculate the vertex, maximum height, and time when the bullet hits the ground, using the equation H(t) = -16t^2 + 736t. Detailed solutions provided, suitable for students in Grades 9-12.

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