Question

A ball was kicked from the ground with a velocity of 64 feet per second and followed the path modeled by the function: h(t)=-16t2+64t. (3 points)

The vertex of the rock's path indicates the rock took

seconds to reach its maximum height of

the graph of this function, we can see the domain in context to this situation is 0 to 4 because the ball

64

2

Iter

Item S

feet. Looking at

Item

hits the ground after 4 seconds.

Item 1

Item 1

Itern 12

C

#reaches a maximum height at 4 feet.

10#4 #48

Answer 1

To analyze the ball's trajectory given by the function \( h(t) = -16t^2 + 64t \), we can determine its vertex, which will tell us the maximum height the ball reaches and the time it takes to reach that height.

### Step 1: Find the Time to Reach Maximum Height
The vertex of a parabola represented by \( h(t) = at^2 + bt + c \) occurs at the time \( t \) given by the formula:
\[
t = -\frac{b}{2a}
\]
In this case, \( a = -16 \) and \( b = 64 \):
\[
t = -\frac{64}{2 \times -16} = -\frac{64}{-32} = 2 \text{ seconds}
\]

### Step 2: Find Maximum Height
Next, we substitute \( t = 2 \) back into the height equation to find the maximum height:
\[
h(2) = -16(2^2) + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]

### Step 3: Determine the Time the Ball Hits the Ground
To find when the ball hits the ground, we need to solve for \( t \) when \( h(t) = 0 \):
\[
-16t^2 + 64t = 0
\]
Factoring out \( 16t \):
\[
16t(-t + 4) = 0
\]
Setting each factor equal to zero gives:
1. \( 16t = 0 \) ⟹ \( t = 0 \) seconds (launch time)
2. \( -t + 4 = 0 \) ⟹ \( t = 4 \) seconds

### Conclusion
The ball reaches its maximum height of **64 feet** at **2 seconds**, and it hits the ground after **4 seconds**.

So, to summarize:
- The ball took **2 seconds** to reach its maximum height of **64 feet**.
- The ball hits the ground after **4 seconds**.