Question

a model rocket is launched with an initial upward velocity of 60 m/s. the rockets height h (in meters) after t seconds is given by the following.
h = 60t - 5t²
find all values of t for which the rockets height is 30 meters.
round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the or button.)

Explanation:

Step1: Substitute \( h = 30 \) into the equation

Given \( h=60t - 5t^{2}\), when \( h = 30\), we have \(30=60t-5t^{2}\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\).
\(5t^{2}-60t + 30=0\). Divide through by \(5\) to simplify: \(t^{2}-12t + 6=0\).
Here \(a = 1\), \(b=-12\), \(c = 6\).

Step2: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 1\), \(b=-12\), \(c = 6\) into the formula:
\(t=\frac{-(-12)\pm\sqrt{(-12)^{2}-4\times1\times6}}{2\times1}=\frac{12\pm\sqrt{144 - 24}}{2}=\frac{12\pm\sqrt{120}}{2}=\frac{12\pm2\sqrt{30}}{2}=6\pm\sqrt{30}\).

Step3: Calculate the values of \(t\)

\(\sqrt{30}\approx5.48\).
For \(t = 6+\sqrt{30}\), \(t\approx6 + 5.48=11.48\).
For \(t=6-\sqrt{30}\), \(t\approx6-5.48 = 0.52\).

Answer:

\(t = 0.52\) or \(t=11.48\)