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.