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Feb 10 2015 09:39pm
My first physics class, basic projectile motion problems.

Free tutoring hosted by our school of engineering -> couldn't solve these problems
Free tutoring hosted by physics department -> couldn't solve these problems
My friend whose majoring in physics -> couldn't solve these problems

Fuck. Any help is appreciated.



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Feb 10 2015 10:04pm
Well for the 1st question u can break the initial velocity into its vertical and horizontal component

horizontal velocity, v_x = v_i * cos (theta)
vertical velocity, v_y = v_i * sin(theta)

neglecting drag, v_x is constant and you know distance so you can find time

d/(v_x) = t

d/ v_i * cos (theta) = t

now use the equation h= v_y * t + (1/2)* at^2 for the vertical component

u know t = d/v_i * cos (theta), a = -g, v_y = v_i * sin(theta)

plug into equation and now you get

h = v_i * sin(theta) * d/v_i * cos (theta)+ (1/2)* (-g) (d/v_i * cos (theta))^2
u can probably simplify but im 2 lazy to try

This post was edited by 2wo1ne on Feb 10 2015 10:06pm
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Feb 10 2015 11:40pm
Quote (2wo1ne @ Feb 10 2015 11:04pm)
Well for the 1st question u can break the initial velocity into its vertical and horizontal component

horizontal velocity, v_x = v_i * cos (theta)
vertical velocity, v_y = v_i * sin(theta)

neglecting drag, v_x is constant and you know distance so you can find time

d/(v_x) = t

d/ v_i * cos (theta) = t

now use the equation h= v_y * t + (1/2)* at^2 for the vertical component

u know t = d/v_i * cos (theta), a = -g, v_y = v_i * sin(theta)

plug into equation and now you get

h = v_i * sin(theta) * d/v_i * cos (theta)+ (1/2)* (-g) (d/v_i * cos (theta))^2
u can probably simplify but im 2 lazy to try


Thanks for quick reply, I will check this in the morning but it looks good and your explanation is very simple.
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Feb 11 2015 03:56am
Quote (2wo1ne @ Feb 10 2015 11:04pm)
Well for the 1st question u can break the initial velocity into its vertical and horizontal component

horizontal velocity, v_x = v_i * cos (theta)
vertical velocity, v_y = v_i * sin(theta)

neglecting drag, v_x is constant and you know distance so you can find time

d/(v_x) = t

d/ v_i * cos (theta) = t

now use the equation h= v_y * t + (1/2)* at^2 for the vertical component

u know t = d/v_i * cos (theta), a = -g, v_y = v_i * sin(theta)

plug into equation and now you get

h = v_i * sin(theta) * d/v_i * cos (theta)+ (1/2)* (-g) (d/v_i * cos (theta))^2
u can probably simplify but im 2 lazy to try


this works, or you can just find the time from t = d/v_i * cos (theta)

then use that time to solve in your height equation h= v_y * t + (1/2)* at^2

If h is positive (congrats, you hit the building, if not, you fell short.
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Feb 11 2015 01:22pm
Quote (TheStealthTarget @ Feb 11 2015 05:56am)
this works, or you can just find the time from t = d/v_i * cos (theta)

then use that time to solve in your height equation h= v_y * t + (1/2)* at^2

If h is positive (congrats, you hit the building, if not, you fell short.


This is the same except u didnt sub in for variables
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Feb 11 2015 04:01pm
Quote (2wo1ne @ Feb 11 2015 02:22pm)
This is the same except u didnt sub in for variables


i know... but it is another way to think about it, and you didn't say anything about how height being negative means it hit the ground before the building.

Just another way to think about it, without 1 large equation
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