Projectile motion - equations giving different answers?

  • Thread starter lilyxmo
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  • #1
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Homework Statement



If an object is being launched from the top of a staircase 1.31m high from a pipe 16cm long, at an angle of 60 degrees. Find where the object will land if the initial velocity is 5.442m/s.

Homework Equations


dx = 2v1^2sinθcosθ / g

dy = vyΔt+1/2gΔt^2

dx = vxΔt


The Attempt at a Solution


I found the components to be
vx = 2.721
vy = 4.713

then i just used these in the two methods
the answers should both be the same, but the dx i got from both was very different
Am i doing something wrong? which one would be the correct answer?



Also, for the first method, i put 120 degrees as negative as it is below my reference point, or it would give me a neg. distance.
 

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Answers and Replies

  • #2
rcgldr
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There is an issue with the wording of the problem. Is the 1.3 m the exit point of the pipe or the base of the pipe? If it's the base, then you'd need to take into account the pipe's 16 cm length. If it's the exit point you wouldn't need to know the length of the pipe, so it seems more likely that 1.3 m is the base of the pipe.
 
  • #3
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to the base of the pipe
so the total height would be the stairs + height of top of stairs to the opening of the pipe

which is 1.324m
 

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  • #4
rcgldr
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Assuming 60 degrees is from horizontal as you've shown, it looks OK so far.
 

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