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April 25, 2024, 03:41:15 am

Author Topic: Projectile Motion Problem  (Read 5243 times)  Share 

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laijiawen

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Re: Projectile Motion Problem
« Reply #30 on: January 05, 2011, 07:30:59 pm »
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surely someone could solve this??? I really need to solve this, so I can have my work completed as I am a perfectionist and need everything done before I move on. =[

bomb

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Re: Projectile Motion Problem
« Reply #31 on: January 05, 2011, 07:35:41 pm »
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surely someone could solve this??? I really need to solve this, so I can have my work completed as I am a perfectionist and need everything done before I move on. =[

You're a perfectionist..for now :P So was I, it lasts like two weeks.
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evaever

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Re: Projectile Motion Problem
« Reply #32 on: January 05, 2011, 07:53:28 pm »
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vertical: u=vtan10, a=-10, s=-1.2, t?   s=ut+0.5at2

horizontal: u=v, s=11.6, t?   s=ut

eliminate t to find v=14.5
so is this the answer?

definitely

evaever

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Re: Projectile Motion Problem
« Reply #33 on: January 05, 2011, 07:54:50 pm »
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No, because like me ha has put distance as 1.2m, when the suitcase goes up first...

-1.2 is the displacement, not the distance

evaever

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Re: Projectile Motion Problem
« Reply #34 on: January 05, 2011, 07:59:32 pm »
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How did you get the initial vertical velocity u= vtan10?

The horizontal component of the velocity vector is v, the vertical component of the velocity vector must be vtan10, so that the velocity vector makes a 10 deg angle. Draw a right triangle to see it.

kamil9876

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Re: Projectile Motion Problem
« Reply #35 on: January 05, 2011, 08:06:54 pm »
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evaever is right about the displacement, I've seen people here being too obsessed with splitting the projectile motion part into the "going up" and "going down" phase, the formula  treats as the displacement over time , so we don't really need to "split up" and can just set s=-1.2 as we take upwards direction as positive.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."