# 3.5 Free fall  (Page 4/7)

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## Solution

1. From [link] , ${v}^{2}={v}_{0}^{2}-2g\left(y-{y}_{0}\right)$ . With $v=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{0}=0$ , we can solve for y :
$y=\frac{{v}_{0}^{2}}{-2g}=\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\text{m}\text{/}{\text{s}\right)}^{2}}{-2\left(9.8\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)}=2040.8\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
2. An altitude of 6.0 km corresponds to $y=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}$ in the coordinate system we are using. The other initial conditions are ${y}_{0}=0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{v}_{0}=200.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ .
We have, from [link] ,
${v}^{2}={\left(200.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\right)}^{2}-2\left(9.8\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}\right)⇒v=±142.8\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}.$

## Significance

We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = −142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.

Visit this site to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

## Summary

• An object in free fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration g due to gravity, which averages $g=9.81\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ .
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, and acceleration.

## Conceptual questions

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? Assume there is no air resistance.

An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have the same sign on the way up as on the way down?

a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes

Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?

Earth $v={v}_{0}-gt=\text{−}gt$ ; Moon ${v}^{\prime }=\frac{g}{6}{t}^{\prime }\phantom{\rule{0.5em}{0ex}}v={v}^{\prime }\phantom{\rule{0.5em}{0ex}}-gt=-\frac{g}{6}{t}^{\prime }\phantom{\rule{0.5em}{0ex}}{t}^{\prime }=6t$ ; Earth $y=-\frac{1}{2}g{t}^{2}$ Moon ${y}^{\prime }=-\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{g}{6}{\left(6t\right)}^{2}=-\frac{1}{2}g6{t}^{2}=-6\left(\frac{1}{2}g{t}^{2}\right)=-6y$

How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?

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