If a ball is thrown into the air at 64 feet per second from the top of a 112​-foot-tall ​building, its height can be modeled by the function Sequals112plus64tminus16tsquared​, where S is in feet and t is in seconds.

Question

If a ball is thrown into the air at 64 feet per second from the top of a 112​-foot-tall ​building, its height can be modeled by the function Sequals112plus64tminus16tsquared​, where S is in feet and t is in seconds.

Answer 1

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

1. 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.

2. 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.

3. 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.