Example Card: Application for an Object in Motion

Let's see how the quadratic formula can model the path of a thrown object from a great height. This example will cover the key idea of applying the quadratic formula to real world scenarios.

Example Problem From an initial height $s$ of 60 meters on a tower, Maria throws a ball up at an initial velocity $v$ of 8 meters per second. Use the equation $ 4.9t^2 + vt + s = 0$ to find the time $t$ in seconds when the ball hits the ground.

Step by Step 1. Substitute the given values into the quadratic equation. Here, $v=8$ and $s=60$. $$ 4.9t^2 + 8t + 60 = 0 $$ 2. We can now use the quadratic formula with $a = 4.9$, $b = 8$, and $c = 60$. $$ t = \frac{ b \pm \sqrt{b^2 4ac}}{2a} $$ 3. Substitute the values into the formula. $$ t = \frac{ (8) \pm \sqrt{(8)^2 4( 4.9)(60)}}{2( 4.9)} $$ 4. Simplify the expression. $$ t = \frac{ 8 \pm \sqrt{64 + 1176}}{ 9.8} = \frac{ 8 \pm \sqrt{1240}}{ 9.8} $$ 5. Since $1240$ is not a perfect square, we find the approximate decimal value. $$ t \approx \frac{ 8 \pm 35.2136}{ 9.8} $$ 6. This gives two possible solutions for time $t$. $$ t \approx \frac{ 8 + 35.2136}{ 9.8} \approx \frac{27.2136}{ 9.8} \approx 2.7769 \text{ seconds} $$ $$ t \approx \frac{ 8 35.2136}{ 9.8} \approx \frac{ 43.2136}{ 9.8} \approx 4.4096 \text{ seconds} $$ 7. Since time cannot be negative, we select the positive solution. The ball will hit the ground in approximately $4.4096$ seconds.