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Motion in 2 Dimensions

  • Thread starter wowdusk
  • Start date
1. Homework Statement
A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal. It's initial speed is 15 m / s. If the brick is in flight for 3 seconds, how tall is the building? Thanks for the help.


2. Homework Equations



3. The Attempt at a Solution
i thought i could use delta y = (Vo*sin*25 degrees)t - 1/2gt^2
but when i plugged everything in i got 63.1m, which is different from my friends 25m. I thought i could use this equation because it seems i have every piece of it and i just needed to plug it in. Did i just find how high the brick was from the ground? I don't know how to find the height of building...i'm confused
 
Last edited:

PhanthomJay

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Your equation is good! Your math is not so good. Recheck your numbers.
 
i still cannot find out what i am doing wrong. My Vo is 15m/s, sin*25 degrees, time is 3 sec, g is -9.8m/s^2

i get the same answer

Whoops...i had g=-9.8, and it had to be a positive in order for this to work.

Thank you!
 
Last edited:

PhanthomJay

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Homework Helper
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i still cannot find out what i am doing wrong. My Vo is 15m/s, sin*25 degrees, time is 3 sec, g is -9.8m/s^2

i get the same answer
You are getting mixed up on your plus and minus signs. If up is positive, then down is negative. I thought you had already built your minus sign into the equation, which should read [tex]y = v_{yi}t +1/2(g)t^2[/tex]. Then plug in g =-9.8. and y comes out to -25, indicating the displacement after 3 seconds is 25 m downward from the top of the building.
 

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