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 need to analyze the given height equation of the ball:

**Step 1: Understand the height equation**
The height \( h(t) \) of the ball at any time \( t \) is given by the equation:
\[ h(t) = -16t^2 + 256t \]
This is a quadratic equation in the standard form \( h(t) = at^2 + bt + c \), where \( a = -16 \), \( b = 256 \), and \( c = 0 \).

**Step 2: Find the velocity**
The velocity \( v(t) \) of the ball is the derivative of the height function with respect to time \( t \):
\[ v(t) = \frac{dh(t)}{dt} = \frac{d}{dt}(-16t^2 + 256t) \]
Calculating the derivative:
\[ v(t) = -32t + 256 \]

**Step 3: Set the velocity to zero**
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 at \( t = 8 \) seconds.

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