Question

as part of his science experiment, burke launched a rocket with an upward velocity of 12 feet per second off a platform which was 3 feet from the ground. the formula in the box describes this experiment.
$h=-4.9t^{2}+12t + 3$
which is closest to the time it will take burkes rocket to hit the ground $(h = 0)$?
2.0 seconds
2.7 seconds
3.0 seconds
3.4 seconds

Explanation:

Step1: Substitute \(h = 0\) into the equation

We have the quadratic equation \(0=-4.9t^{2}+12t + 3\). For a quadratic equation \(ax^{2}+bx + c=0\) (here \(a=-4.9\), \(b = 12\), \(c = 3\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant

First, calculate the discriminant \(\Delta=b^{2}-4ac=(12)^{2}-4\times(-4.9)\times3=144 + 58.8=202.8\).

Step3: Calculate \(t\) using the quadratic formula

\(t=\frac{-12\pm\sqrt{202.8}}{2\times(-4.9)}\). We consider the positive root (since time \(t>0\)). \(t=\frac{-12+\sqrt{202.8}}{-9.8}\approx\frac{-12 + 14.24}{-9.8}\) (because \(\sqrt{202.8}\approx14.24\)). \(t=\frac{2.24}{-9.8}\) (wrong, should be \(t=\frac{- 12-\sqrt{202.8}}{2\times(-4.9)}=\frac{-12 - 14.24}{-9.8}=\frac{-26.24}{-9.8}\approx2.7\)).

Another way:

Step1: Use substitution

We can also test the values.
For \(t = 2.0\): \(h=-4.9\times(2)^{2}+12\times2 + 3=-19.6+24 + 3=7.4\).
For \(t = 2.7\): \(h=-4.9\times(2.7)^{2}+12\times2.7+3=-4.9\times7.29+32.4 + 3=-35.721+32.4+3\approx - 0.321\approx0\).
For \(t = 3.0\): \(h=-4.9\times(3)^{2}+12\times3+3=-44.1+36 + 3=-5.1\).
For \(t = 3.4\): \(h=-4.9\times(3.4)^{2}+12\times3.4+3=-4.9\times11.56+40.8+3=-56.644+40.8+3=-12.844\).

Answer:

2.7 seconds