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Projectile's launch speed?

  • Thread starter kcalhoun
  • Start date
1. 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?
2. Homework Equations
kinematics equations


3. 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
 

LowlyPion

Homework Helper
3,079
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What does clearing a maximum height barrier suggest?

Will that be the highest point of its trajectory?
 
yes 15m is the highest point i just don't know how to use that to solve the problem
 

LowlyPion

Homework Helper
3,079
4
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?
 
vsin13?
 
i don't know how to find Vo with the given info.. thats my problem
 
so use v^2=Vo^2+2aX?? But i still dont know what v or Vo are or how to get them...
 

LowlyPion

Homework Helper
3,079
4
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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