Projectile's launch speed?

  • Thread starter kcalhoun
  • Start date
  • #1
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Homework Statement



The highest barrier that a projectile can clear is 15.0 m, when the projectile is launched at an angle of 13.0° above the horizontal. What is the projectile's launch speed?

Homework Equations


kinematics equations


The Attempt at a Solution



delta y=15m v0x= vcos13
ay=-9.8m/s^2 ax=0m/s^2

im not sure how to use the information that i have to solve the problem
 

Answers and Replies

  • #2
LowlyPion
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What does clearing a maximum height barrier suggest?

Will that be the highest point of its trajectory?
 
  • #3
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yes 15m is the highest point i just don't know how to use that to solve the problem
 
  • #4
LowlyPion
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yes 15m is the highest point i just don't know how to use that to solve the problem
What will the vertical component of velocity need to be if it is only to go 15 m high?
 
  • #5
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vsin13?
 
  • #6
LowlyPion
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vsin13?
Yes it will equal that.

But they ask you to find Vo to begin with.
 
  • #7
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i don't know how to find Vo with the given info.. thats my problem
 
  • #9
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so use v^2=Vo^2+2aX?? But i still dont know what v or Vo are or how to get them...
 
  • #10
LowlyPion
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so use v^2=Vo^2+2aX?? But i still dont know what v or Vo are or how to get them...
So at max height, what will the vertical velocity be again?
 

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