Can a Bowling Ball Really Travel 200 Meters After Launching from a Ramp?

[kurgen88]

Homework Statement

A solid bowling ball rolls down a hill with a height of 12.00 meters. The end of the hill has a ramp/jump which has an unknown angle above the horizontal. When the ball leaves the ramp at the unkown angle, its speed is13.09 meters per second and travels a distance of 200.0 meters.

I need to solve for the angle which the ball leaves the ramp, but the fact that the ball travels 200.0 meters seems like it is impossible. Is this a trick question? Does anyone have any suggestions on how I can solve for this?

Homework Equations

gravity*distance=-2V^2 * sin(theta)cos(theta)

or...

-Sin(2theta) = (gravity/initial velocity^2)

...where gravity = 10.0 meters/sec^2

I have tried to solve this way but I don't think I can isolate theta from Sin(2

 

[andrevdh]

Look in your handbook at the section on projectile motion. Try and find the range equation of a projectile. The range of a projectile is the horizontal distance it travels from the launching point up to the point where it hits the ground.

 

[kurgen88]

I am at the point given in the equation below. Is it possible to solve for theta here? Can you use algebra to extract theta form sin(2*theta) ? What happens to sin(2 ?


-Sin(2theta) = (gravity/initial velocity^2)

 

[andrevdh]

If you take the inverse sine of the calculated value (arcsin) you will get an angle. This angle is then twice the angle you want (2\theta _o).

Note that in your equation above you omitted R.