#1
An object is thrown in the air from an initial height of 12 feet, with an initial upward velocity of 16 feet/second?

* How long will the object be in the air?
* What will the velocity of the object be after 1 second?

I don't even know where to start. Is there a formula i'm forgetting? Please help.
#3
This is physics you a*s.

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#4
gravitational force is 9.81m/s^2 so you probly need that in f/s^2 so convert and figure it out
#5
ahhh physics!
you're gonna need the angle that it's thrown at...
just remember that the acceleration of gravity is 9.8 meters per second squared.... convert your feet to meters... and then look up the equation for a parabola cuz i dont remember it off the top of my head
and plug and chug, baby!
#6
Quote by barden1069
ahhh physics!
you're gonna need the angle that it's thrown at...
just remember that the acceleration of gravity is 9.8 meters per second squared.... convert your feet to meters... and then look up the equation for a parabola cuz i dont remember it off the top of my head
and plug and chug, baby!



hey friend sorry but he said straight up not on an angle and im sure its just intro to physics therefore perfect vacuum is in effect, therefore no air friction so there is no vectors in this problem.
#8
Quote by barden1069
ahhh physics!
you're gonna need the angle that it's thrown at...
just remember that the acceleration of gravity is 9.8 meters per second squared.... convert your feet to meters... and then look up the equation for a parabola cuz i dont remember it off the top of my head
and plug and chug, baby!


i think i've got it:

h(t) = -16t^2 + vt + h (in feet)

if that works, i totally owe you a cookie. btw this is the 1st problem of an online calculus assignment due in just over an hour, so i'm sort of crapping myself

EDIT: i'm fairly sure thats the formula for vertical motion, now i think about it, but you reminded me of it somehow
Last edited by H(i)SS at Jan 13, 2009,
#9
Quote by H(i)SS
An object is thrown in the air from an initial height of 12 feet, with an initial upward velocity of 16 feet/second?

* How long will the object be in the air?
* What will the velocity of the object be after 1 second?

I don't even know where to start. Is there a formula i'm forgetting? Please help.


For #1
This equation

distance=v(initial)+1/2*accel*time^2
or
d = vi+1/2at^2

Same thing

First off, a = 9.8, or acceleration due to gravity.

To find the time spent in the air, first find the time it spends going up and then back down to the same height, then add the time it then takes to hit the ground

For #2

acceleration=change in velocity/time
a=dv/t

Acceleration is 9.8 gravity again. Your time is 1 second. Your change in velocity can now be found.

Tada!
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#10
I took physics last year and do not have the formulas memorized, but I will tell you how to find the answer. The angle does not matter. You need to use an equation that uses initial velocity and final velocity to determine how long it takes for the object to reach 0 f/s while traveling against gravity. Gravity is 32 f/s so I think you would use -32 f/s in this problem. Once you know how high it went before it stopped, you can then calculate how long it will take to fall to the ground since all objects fall at the same speed (as long as we are not taking air resistance into account, which in this case it doesn't matter). That is how you do the first part.

For the second part, I think you need to find a formula that incorporates acceleration, time, velocity, and maybe distance? Then solve for t time = 1.

Again, I took this class last year as a junior and am not taking a science class my senior year so I am not 100% certain. However, I think this should help get you started. Good luck!
#11
First of all, why the hell are you using feet?

I thought Physics was all about MKS (meters, kilos, seconds).

I would help you, but I hate converting feet to meters, it makes everything a pain in the ass
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#12
Sorry if I'm late or beaten, but Google Kinematic Equations.
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#13
Quote by algemar
For #1
This equation

distance=v(initial)+1/2*accel*time^2
or
d = vi+1/2at^2

Same thing

First off, a = 9.8, or acceleration due to gravity.

To find the time spent in the air, first find the time it spends going up and then back down to the same height, then add the time it then takes to hit the ground

For #2

acceleration=change in velocity/time
a=dv/t

Acceleration is 9.8 gravity again. Your time is 1 second. Your change in velocity can now be found.

Tada!


You are switching between feet and meters in your distances.

TS, just convert everything you have in feet to meters, and then do this. It looks about right.
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#14
thanks for the help guys, but i got it! that vertical motion equation (in feet) is the right one.

The reason everything is not in meters is because it's not actually physics or any science for that matter... it's calculus, and those are the units i have
#15
Quote by H(i)SS
thanks for the help guys, but i got it! that vertical motion equation (in feet) is the right one.

The reason everything is not in meters is because it's not actually physics or any science for that matter... it's calculus, and those are the units i have


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#17
Quote by H(i)SS
idk about all that, but i just finished the assignment...


One sec...
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#18
Quote by zippidyduda
First of all, why the hell are you using feet?

I thought Physics was all about MKS (meters, kilos, seconds).

I would help you, but I hate converting feet to meters, it makes everything a pain in the ass


Converting is hardly a pain.

There's always the FPS (foot pound second) systems, absolute and gravitational (I'm an absolute sorta guy - I likes the poundal).
#19
Quote by H(i)SS
idk about all that, but i just finished the assignment...



I FOUND IT!
You can call me Aaron.


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Out on parole, any more instances of plum text and I get put back in...