The height (in meters) at any time t (in seconds) of a ball thrown vertically varies according to equation `h(t)=-16t^(2)+256t`. How long after in seconds the ball reaches the hightest point

Answer

To find out how long after the ball reaches the highest point, we will analyze the given height equation and determine when the velocity of the ball becomes zero. The height of the ball at any time \( t \) is given by the equation:

\[ h(t) = -16t^2 + 256t \]

### Step 1: Differentiate the height equation to find the velocity
The velocity \( v(t) \) is the derivative of the height function \( h(t) \) with respect to time \( t \).

\[ v(t) = \frac{dh}{dt} = \frac{d}{dt}(-16t^2 + 256t) \]

### Step 2: Apply the power rule to differentiate
Using the power rule for differentiation, we differentiate each term:

\[ v(t) = -16 \cdot 2t + 256 \cdot 1 = -32t + 256 \]

### Step 3: Set the velocity equal to zero to find the time at the highest point
At the highest point, the velocity of the ball is zero. Therefore, we set the velocity equation to zero:

\[ -32t + 256 = 0 \]

### Step 4: Solve for \( t \)
Now, we solve for \( t \):

\[ -32t = -256 \]
\[ t = \frac{256}{32} \]
\[ t = 8 \]

### Conclusion
The ball reaches its highest point after \( t = 8 \) seconds.

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