What Is the Launch Speed for a Projectile to Clear a 15m Barrier?

[kcalhoun]

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

 

[LowlyPion]

What does clearing a maximum height barrier suggest?

Will that be the highest point of its trajectory?

 

[kcalhoun]

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

 

[LowlyPion]

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?

 

[kcalhoun]

vsin13?

 

[LowlyPion]

vsin13?

Yes it will equal that.

But they ask you to find Vo to begin with.

 

[kcalhoun]

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

 

[LowlyPion]

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

Here are some formulas for kinematic equations.
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Your a = g here.

So all you need is a formula to relate Velocity to distance and acceleration.

 

[kcalhoun]

so use v^2=Vo^2+2aX?? But i still don't know what v or Vo are or how to get them...

 

[LowlyPion]

so use v^2=Vo^2+2aX?? But i still don't know what v or Vo are or how to get them...

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