# Boy Scouts are firing model rockets in a field. One boy used a radar gun to determine his rocket had a lift-off velocity of 34 m/s. How long did it take for the rocket to return to the ground? Assume the rocket went straight up and came straight down.

Aug 30, 2015

It took the rocket 6.94 seconds to return back to ground level.

#### Explanation:

So, the Boy Scouts are firing model rockets in a field.

It is important to note here that the rockets are being launched straight up, which means that their trajectory will be restricted to the vertical axis.

The idea here is that while it's moving straight up into the air, the rocket is being decelerated by the gravitational acceleration, $g$, which acts towards the ground.

This means that the rocket will continue to go up until it reaches maximum height, at which its velocity is equal to zero.

From that point on, it will free fall towards the ground, this time being accelerated by the gravitational acceleration.

So, if its velocity is zero at maximum height, you can use its initial velocity to figure out how much time it needs to reach that height

underbrace(v_"top")_(color(blue)(=0)) = v_0 - g * t_"up" implies t_"up" = v_0/g

${t}_{\text{up" = (34color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "3.47 s}}$

So, the rocket climbed for 3.47 seconds. From this point on, the rocket will start its free fall motion. Since the gravitational acceleration wwill now accelerate the rocket over the same distance, the time ittakes the rocket to free fall will be equal to the time it took it to reach maximum height.

Therefore, the total time of flight will be

t_"total" = 2 * t_"up" = 2 * 3.47 = color(green)("6.94 s")

Alternatively

You can calculate the time needed for the rocket to reach ground from maximum height by calculating maximum height.

${\underbrace{{v}_{\text{top}}^{2}}}_{\textcolor{b l u e}{= 0}} = {v}_{0}^{2} - 2 \cdot g \cdot h \implies h = {v}_{0}^{2} / \left(2 \cdot g\right)$

h = (34^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = "59.0 m"

Now you just need to calculate how much time it takes the rocket to free fall from $h = \text{59.0 m}$. This time ${v}_{\text{top}}$ will act as the rocket's initial speed for the free fall period of its flight.

h = underbrace(v_"top")_(color(blue)(=0)) * t + 1/2 * g * t_"fall"^2

${t}_{\text{fall" = sqrt((2 * h)/g) = sqrt( (2 * 59.0color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^2)) = "3.47 s}}$

The total time it takes the rocket to return to ground level is

${t}_{\text{total" = t_"up" + t_"down}}$

t_"total" = 3.47 + 3.50 = color(green)("6.94 s")