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.
