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?
equation: $-5t^2 + 13t - 6 = 0$

Explanation:

Step1: Identify the quadratic equation

We have the quadratic equation $-5t^{2}+13t - 6=0$. Multiply both sides by - 1 to make the coefficient of $t^{2}$ positive: $5t^{2}-13t + 6 = 0$.

Step2: Use the quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$, the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 5$, $b=- 13$, $c = 6$.

First, calculate the discriminant $\Delta=b^{2}-4ac=(-13)^{2}-4\times5\times6=169 - 120 = 49$.

Then, $t=\frac{13\pm\sqrt{49}}{2\times5}=\frac{13\pm7}{10}$.

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.