The height h , in feet , of a ball shot upward from a ground level spring gun is described by the formula `h=-16t^2+48t`. Where t is the time in seconds. What is the maximum height , in feet , reached by the ball ?

Answer

To find the maximum height reached by the ball, we will follow these steps:

### Step 1: Identify the height function
The height \( h \) of the ball is given by the equation:
\[ h = -16t^2 + 48t \]

### Step 2: Differentiate the height function
To find the time at which the maximum height occurs, we need to differentiate the height function with respect to time \( t \):
\[ \frac{dh}{dt} = \frac{d}{dt}(-16t^2 + 48t) \]
Using the power rule, we get:
\[ \frac{dh}{dt} = -32t + 48 \]

### Step 3: Set the derivative equal to zero
To find the critical points where the maximum height occurs, we set the derivative equal to zero:
\[ -32t + 48 = 0 \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:
\[ 32t = 48 \]
Dividing both sides by 32, we find:
\[ t = \frac{48}{32} = \frac{3}{2} \text{ seconds} \]

### Step 5: Substitute \( t \) back into the height function
Now we will substitute \( t = \frac{3}{2} \) back into the height function to find the maximum height:
\[ h = -16\left(\frac{3}{2}\right)^2 + 48\left(\frac{3}{2}\right) \]
Calculating \( \left(\frac{3}{2}\right)^2 \):
\[ \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
Now substituting this value into the height equation:
\[ h = -16\left(\frac{9}{4}\right) + 48\left(\frac{3}{2}\right) \]
Calculating each term:
\[ h = -36 + 72 \]
Finally, we find:
\[ h = 36 \text{ feet} \]

### Conclusion
The maximum height reached by the ball is:
\[ \boxed{36 \text{ feet}} \]