Question

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

Answer 1

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.