Question

An astronaut drops a feather from 4.8 m
above the surface of the moon.
If the acceleration of gravity on the moon
is 1.65 m/s
2
, how long does it take the feather
to reach the surface?

Answer 1

Hello! You want to look at what you know!
Distance to the ground?
Gravitational acceleration?

Now, what else?
Initial velocity. See, when a problem says that you "drop" something, it assumes that you were holding it still, more or less. So, initial velocity is 0.

Now you need to know or derive a formula that uses your known information to find the time it takes the feather to fall.

A common way is to use one of the fundamental kinematic equations, which I adapt:
\(\Delta x=v_0t+\frac12at^2\)
\(\downarrow\)
\(\Delta y=(v_{initial})t+\frac12gt^2\)

From there, you should know where to go if you understand why we did that. The formula needed to have \(t\) and some of what we know (\(\Delta y, g, v_{initial}\)).

We want to solve for \(t\). Note that you can start by making \(v_{initial}=0\) before solving for \(t\). That will be easier.

Answer 2

i dont get it?????

Answer 3

Hi! Where did it start to get fuzzy? You can even say "the beginning" if you want.

Answer 4

the beginning

Answer 5

Okay!
So, you understand that this astronaut is dropping a feather on the moon.

The first thing we need to do is look for \(facts\) that we can use in a mathematical scenario! Pretty much, identify some variables that we hope to use.

I haven't tried this before, but why don't you describe what's happening in this scenario, and then I'll annotate it with physics jargon!

Answer 6

This way you'll see where all of the physics comes in, or I can point you to anything you've missed.

Answer 7

Like, play-by-play, one or two sentences maybe. Or I can just go through it on my own. Your choice!

Answer 8

Oh, you're offline!..

Answer 9

um play by play

Answer 10

"how long does it take the feather to reach the surface?"

An astronaut drops a feather
\(\color{red}{\bullet}\) The feather is dropped, but it was held still just before that.
\(\quad\)\(\color{#1111CC}{\rm Initial\ velocity}\) is \(\color{#11BB11}{0\rm\ m/s}\).

from \(4.8\ \rm m\) above the surface of the moon.
\(\color{red}{\bullet}\) The height of the fall, which is the \(\color{#1111CC}{\rm change\ in\ position}\), is \(\color{#11BB11}{4.8\rm\ m}\).

If the acceleration of gravity on the moon is \(1.65\ \rm m/s^2\) ,
\(\color{red}{\bullet}\) Gravitational acceleration is \(1.65\rm\ m/s^2\), so the feather will \(\color{#1111CC}{\rm accelerate}\) at \(\color{#11BB11}{1.65\rm\ m/s^2}\).

how long does it take the feather to reach the surface?

Just so we can talk about these numbers in math, we give these quantities variable names, \(\color{#1111CC}{v_i,\ \Delta y,\ g}\).

Knowing the correct equation then becomes the key, along with the knowledge that we're looking for the time, \(\color{#FF0000}{t}\).

So, since you're at a loss, I provide you with this equation:
\(\Large\color{#1111CC}{\Delta y}=\color{#1111CC}{v_i}\color{#FF0000}t +\frac12\color{#1111CC}{g}\color{#FF0000}t^2\)

So, you must solve for \(\color{red}t\). But putting in the value for \(\color{blue}{v_i}\) to start will help you do that more easily.

Good luck!

Answer 11

thank you

Answer 12

You're welcome! :) Feel free to ask anything else!