The height \( S \) of the ball at time \( t \) seconds is modeled by the function:
\[ S(t) = 112 + 64t - 16t^2 \]
Where:
- \( S(t) \) is the height of the ball in feet,
- \( t \) is the time in seconds,
- The ball starts 112 feet above the ground,
- It has an initial velocity of 64 feet per second,
- The term \( -16t^2 \) represents the effect of gravity on the ball, causing it to decelerate.
Key Points of the Function
-
Finding the Height at Specific Times:
- You can substitute different values of \( t \) into the equation to find out how high the ball is at that specific time.
-
Maximum Height:
- To find the time at which the ball reaches its maximum height, you can use the vertex formula for a quadratic function. The time at which it reaches its maximum height \( t \) can be found using: \[ t = -\frac{b}{2a} \] where \( a = -16 \) and \( b = 64 \): \[ t = -\frac{64}{2 \cdot -16} = 2 \text{ seconds} \] Then, substitute \( t = 2 \) back into the height function to find the maximum height: \[ S(2) = 112 + 64(2) - 16(2^2) \] \[ S(2) = 112 + 128 - 64 = 176 \text{ feet} \] So the maximum height of the ball is 176 feet.
-
Finding When the Ball Hits the Ground:
- To find when the ball hits the ground, set \( S(t) = 0 \): \[ 0 = 112 + 64t - 16t^2 \] Rearranging gives: \[ 16t^2 - 64t - 112 = 0 \] Dividing the entire equation by 16: \[ t^2 - 4t - 7 = 0 \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \] \[ t = \frac{4 \pm \sqrt{16 + 28}}{2} = \frac{4 \pm \sqrt{44}}{2} \] \[ t = \frac{4 \pm 2\sqrt{11}}{2} = 2 \pm \sqrt{11} \] Since time cannot be negative, we take the positive value: \[ t = 2 + \sqrt{11} \approx 5.32 \text{ seconds} \]
Summary
- The maximum height of the ball is 176 feet, reached at 2 seconds.
- The ball hits the ground approximately 5.32 seconds after being thrown.
4 answers