# 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. When the rocket was traveling at 17m/s, how far was it from its highest point?

Aug 31, 2015

It was 14.7 m from its maximum height.

#### Explanation:

So, you know that the rockets are being launched with an initial velocity of 34 m/s.

Since the rocket is moving straight up, it will be decelerated by the gravitational acceleration, $g$. This means that its velocity will continue to decrease until it reaches maximum height, where it will be equal to zero.

This means that you can determine the maximum height reached by the rocket by using the formula

${\underbrace{{v}^{2}}}_{\textcolor{b l u e}{= 0}} = {v}_{0}^{2} - 2 \cdot g \cdot h$

Rearrange to solve for $h$

h = v_0^2/(2 * g) = (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"

Since on its way up the rocket is being decelerated by $g$, and on its way down it's being accelerated by $g$, that can only mean that it will reach the ground having the same velocity as it did when it was launched.

The distance is the same, so it goes from 34 m/s to 0 m/s on its way up, and from 0 m/s to 34 m/s on its way down.

Now, after the rocket reaches maximum height, it begins to free fall towards the ground. Since at the top of its trajectory its velocity is equal to zero, you can use the same formula to determine the distance ${h}_{\text{down}}$ it travelled until its velocity reached 17 m/s.

v^2 = underbrace(v_"top"^2)_(color(blue)(=0)) + 2 * g * h_"down"

${h}_{\text{down" = v^2/(2 * g) = (17^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = "14.7 m}}$

Therefore, the rocket is 14.7 meters from its maximum height when it has a velocity of 17 m/s.

Alternatively

You can calculate the distance from the top of the rocket's trajectory by calculating the distance it covered on its way up until its velocity reached 17 m/s, then subtracting this value from the max height.

${17}^{2} = {34}^{2} - 2 \cdot g \cdot {h}_{\text{up}}$

${h}_{\text{up" = (34^2 - 17^2)/(2 * 9.8) = "44.2 m}}$

This means that it is

$59.0 - 44.2 = \text{14.8 m}$

from the top - the difference in the value comes from rounding along the way.