Calculating Horizontal Distance of Ball Bearing with Mass 11.0 g

[SMS]

A small steel ball bearing with a mass of 11.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.37 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 30.0 deg from the horizontal.

I started off by getting the initial velocity for when it was released vertically and got vo=5.18m/s.

Then I used the initial velocity to find time in the air using cos theta and got t=.08s but I do not think that is right and this is where I got stuck.

So if someone can show me what I did wrong or missed it would help.

Thanks,
SMS

 

[Enginator]

The energy expended by the spring to get the ball to that height will be the same energy in shooting it horizontally.

 

[faust9]

You don't need to calculate the time. You found the V_o now simply stick that into the projectile range equation. If you don't know this off hand you can derive it by solving the 'y' equation for 't' where y=0 (the bearing has landed thus zero) and substituting that equation in terms of 't' into the 'x' equation. It's actually pretty easy to do and if you know your trig identies you'll have a 3 term equation on the right side where 2 terms are constants(the angle and gravity).

Good luck.

 

[Tide]

When projected at 30 degrees exactly half of the ball bearing's initial speed is in the vertical direction which means that \frac {1}{4} of its kinetic energy is available for conversion to potential energy. The bearing will rise to \frac{1}{4} of the original height.

 

[SMS]

Thanks

Thanks to everyone in here who helped. I got the answer after beating myself with a hammer and realizing how easy it was.