Question

quadratic word problems
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 ?

Explanation:

Step1: Substitute values into the formula

We know the height formula is \( h = -5t^2 + vt \), where \( v = 13 \) m/s (initial speed) and \( h = 6 \) m (target height). Substitute these values into the formula:
\( 6 = -5t^2 + 13t \)

Step2: Rearrange into 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 solve the quadratic equation \( 5t^2 - 13t + 6 = 0 \) using the quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = -13 \), \( c = 6 \).
First, calculate the discriminant \( D = b^2 - 4ac = (-13)^2 - 4\times5\times6 = 169 - 120 = 49 \)
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 \( t_2=\frac{13 - 7}{10}=\frac{6}{10}=0.6 \)

Step4: Determine the first time

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.