How do I solve for the release height to land in a hole on a ramp?

[Sean77771]

Homework Statement

A small block is placed at height h on a frictionless, 30 degree ramp. Upon being released, the block slides down the ramp and then falls 1.0m to the floor. A small hole is located 1.0m from the end of the ramp. From what height h shold the block be released in order to land directly in the hole?

Homework Equations

All the basic kinematics formulae.

The Attempt at a Solution

I figured out that the horizontal distance that it travels on the ramp will be 1.73h. I set 1.73h equal to 9.8(h+1) for the distance it will be accelerating in each direction. This came out to be 1.214, which was the wrong answer. Don't know what to try now. There seem to be too many variables to do anything else. Please help. Thanks :)

 

[Kurdt]

That is a rather involved question. If you could show your working perhaps someone would havea chance of spotting where you went wrong.

 

[Sean77771]

Where I went wrong was not knowing how to do it in the first place. I think my solution is completely off base, how would I even begin to solve this?

 

[Kurdt]

You will need to do quite a bit of resolving of forces. The block is on a ramp and so it has a force due to gravity accelerating it parallel to the ramp. You will have to resolve the weight of the block along the ramp to find this force. From that you can find the acceleration. Then you will have to resolve that into x and y components so you can come up with some values for speed at the end of the ramp.

Part two will be when the block leaves the ramp. It will have constant speed in the x direction now and will feel the full acceleration due to gravity in the y direction. It'll be very messy I suspect but will be a good test of your understanding.

 

[Sean77771]

Ok, part two of your description I can get no problem. How do I resolve the weight/find the x and y components as you described in part one of your answer?

 

[Kurdt]

Its a case of some trigonometry. If one draws a free body diagram with the box on the slope, the weight acts vertically downwards. The normal force acts perpendicular to the slope and ther will be a component of weight parallel to the slope. These 3 force vectors act as the sides of a right angled triangle with the weight being the hypotenuse. The normal force is mg cos (theta) and the weight parallel to the slope is mg sin (theta).

Work out what the velocity will be at the end of the ramp in terms of h and then you will have to take the components of this velocity to use in the second part. You will find you have two unknowns and two equations fro x and y components. When I said it seems like it would get messy before it doesn't as I've just tried it.

Remember: sin (30) = 1/2
Remember: cos (30) = (\surd3)/2

Hopefully that should get you started.

 

[Sean77771]

K great thanks for your help