Newton's Laws Problem- Question on upcoming TEST

[mrmiggidude]

Newton's Laws Problem- Question on upcoming TEST!

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 then falls 1.0 m to the floor. A small hole is located 1.0 m from the end of the ramp. From what height h should the block be released in order to land in the hole?

Homework Equations

Newton's Laws
Kinematics Equations

The Attempt at a Solution

I did the problem assuming there would be no initial Y-velocity.
And figured out the time it would take for the object to fall to the ground after leaving the ramp, and the X-velocity it would need to get to the whole.
The time was .45 seconds and the X-velocity was 2.21 m/s.
With that it should be really easy to find the time to get to that velocity, and then the height needed to start from...

But I realized there would be an initial velocity in the y direction when it reaches the end of the ramp. I've spent nearly 3 hours trying to solve this problem, and it might be on an upcoming test. I have tried the problem with different axis, and I just can not figure it out. Please help, my test is this week!

 

[krausr79]

After leaving the plane, the block will have an unknown speed (S), which can be broken into x,y components. The x,y kinematics are:

x - 1
x0 - 0
V - ?
V0 - Scos(-30)
a - 0
t - t

y - 0
y0 - 1
V - ?
V0 - Ssin(-30)
a - -9.8
t - t
(note: time will be the same in both)

In both cases, V is what we don't care about. We then use x=x0+v0t+.5at²
in the x case:
1=0+Scos(-30)+.5(0)t²
Solving for t gives t=1/Scos(-30)

in the y case, with t-substitution:
0=1+Ssin(-30)*1/Scos(-30)+.5(-9.8)(1/Scos(-30))²

In the middle velocity term, S cancels out. Multiplying all known numbers as well as squaring the third term gives:
0=1-.5774-6.5333/S²

Solving for S gives S = 3.9319 m/s

Now with a target final ramp speed, we can set up a 3rd kinematics equation for the slide. The acceleration will be modified from gravity: Slide force = 9.8*sin(30) = 4.9 m/s².
x - ?
x0 - 0
V - 3.9319
V0 - 0
a - 4.9
t - ?
We don't care about time in this one, so the equation we use is V² = V0²+2a(x-x0).
3.9319²=0+2*4.9*(x-0)
Solving for x gives x=1.5775 m up the ramp. They said they wanted height and not ramp distance, so we find the y component of the ramp length: 1.5775*sin(30) = .78875 m 'high' on the ramp.

Brutal!

 

[mrmiggidude]

Yes! That was totally it. Brutal problem, indeed!