What is the velocity of a wrecking ball at a 20.0 degree angle?

[xrotaryguy]

A wrecking ball suspended from a 12.0m cable is accelerated up to 5.00m/s at it's lowest point. After that point the ball has negative aceleration as it gains elevation. What is the ball's velocity when the cable makes a 20.0 degree angle with the verticle?

Is this just a matter of finding the height of the ball at that point and calculating the velocity at that height? I guess I could use the potential energy formula (U=mgh) some how. I'm just not sure on this one.

 

[BerryBoy]

I like the U=mgh idea; you know what the energy of the system is because your given the speed (therefore the kinetic energy) at the lowest point.

 

[AVD]

i don't understand are you saying that its velocity at the lowest point is 5.00m/s ? If so try using conservation of energy.

 

[xrotaryguy]

i don't understand are you saying that its velocity at the lowest point is 5.00m/s ? If so try using conservation of energy.

Oops, in my initial post, I meant to say that the ball was accelerated to 5.00m/s, so yes, the velocity at the lowest point is 5.00m/s.

 

[matthew baird]

Oops, in my initial post, I meant to say that the ball was accelerated to 5.00m/s, so yes, the velocity at the lowest point is 5.00m/s.

OKay, then your on the right track. Use conservation of energy. Set up the equation like this:
KE=KE+mgh so...
1/2mVi^2=1/2mVf^2+mgh, where Vi=initial velocity=5m/s, and Vf=unknown
The mass cancels out, your left with:
1/2Vi^2=1/2Vf^2+gh
use trig to find h at 20degrees, then plug in your values to find final velocity.

 

[xrotaryguy]

Thanks, got it!

V=sqrt(((5.00m/s)^2)-2(9.8)(12.0-12.0cos20.0))=3.29m/s