Question

a rocket is launched straight up at a speed of 13 meters per second. the height of the rocket in meters is $h = -5t^2 + vt$, with $v$ as the initial speed. when will the rocket first reach a height of 6 meters? when will the rocket first reach a height of 6m?

Explanation:

Step1: Substitute values into the formula

We know that \( h = 6 \) meters and \( v = 13 \) m/s. Substitute these into the height formula \( h=-5t^{2}+vt \), we get:
\( 6=-5t^{2}+13t \)

Step2: Rearrange the equation to standard quadratic form

Rearrange the equation to \( ax^{2}+bx + c = 0 \) form (here we use \( t \) instead of \( x \)):
\( 5t^{2}-13t + 6 = 0 \)

Step3: Solve the quadratic equation

We can use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 5 \), \( b=- 13 \), \( c = 6 \). First, calculate the discriminant \( \Delta=b^{2}-4ac=(-13)^{2}-4\times5\times6=169 - 120 = 49 \)
Then, \( t=\frac{13\pm\sqrt{49}}{10}=\frac{13\pm7}{10} \)
We get two solutions: \( t_{1}=\frac{13 + 7}{10}=\frac{20}{10}=2 \) and \( t_{2}=\frac{13-7}{10}=\frac{6}{10}=0.6 \)
Since we want the first time the rocket reaches 6 meters, we take the smaller value of \( t \).

Answer:

The rocket first reaches a height of 6 meters at \( t = 0.6 \) seconds.