Asmall 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 1.0 m from the end of the ramp. from what height h should the block be released in order to land in the hole?

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

We then use x=x0+v0t+.5at^2

in the x case: 1=0+Scos(-30)+.5(0)t^2

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/S^2

Solving for S = S = 3.9319 m/s

Now with a mark on final ramp speed, we can now make a 3rd kinematics equation. The acceleration will be altered from gravity:

Slide force = 9.8*sin(30) = 4.9 m/s^2.

x - ?

x0 – 0

V - 3.9319

V0 – 0

a - 4.9

t - ?

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.

So we now look for the y component of the ramp length:

1.5775*sin(30) = .78875 m 'high' on the ramp.