# Kinematics problem

1. Mar 2, 2009

### kcalhoun

1. The problem statement, all variables and given/known data

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. Relevant 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

2. Mar 2, 2009

### LowlyPion

What does clearing a maximum height barrier suggest?

Will that be the highest point of its trajectory?

3. Mar 2, 2009

### kcalhoun

yes 15m is the highest point i just don't know how to use that to solve the problem

4. Mar 2, 2009

### LowlyPion

What will the vertical component of velocity need to be if it is only to go 15 m high?

5. Mar 2, 2009

vsin13?

6. Mar 2, 2009

### LowlyPion

Yes it will equal that.

But they ask you to find Vo to begin with.

7. Mar 2, 2009

### kcalhoun

i don't know how to find Vo with the given info.. thats my problem

8. Mar 2, 2009

9. Mar 2, 2009

### kcalhoun

so use v^2=Vo^2+2aX?? But i still dont know what v or Vo are or how to get them...

10. Mar 2, 2009

### LowlyPion

So at max height, what will the vertical velocity be again?