How Do You Calculate the Distance a Block Travels Off a Ramp?

[dantra]

Homework Statement

A block is being launch off a ramp, the length of the ramp is 5m(hypotenuse). The ramp also has a kinetic coefficient of .2 and the block has an initial velocity of 12m/s.

Homework Equations

-Wx-uFn = ma

Velocity equations for finding distance off the ramp

The Attempt at a Solution

ok to find the velocity at the end of the ramp.

-mgsin35-umgcos(35) = ma
15.6 - 1.6 = -7.2 s

V^2 + Vi^2 + 2at(change in x)
V= Square root(12^2 + 2(-7.2)(5))
V at the end of the ramp is 8.47 ms (this was confirm right by the answer sheets)

finding the velocity in all direction

Vy = 4.86 ms
Vx = 6.94 ms
V = 8.47 ms

now for the distance, I am going to find the time it takes for Vy to reach zero

Vyf = Vyi +at
0 = 4.8 + (-9.8)t
t=.49

time it takes for it to fall down
Height of the ramp 2.87
Height of the pojectable when Vy = 0 2.384
total height 5.25

0= 5.25 + 1/2(-9.8)t
T= 1.07s

Total time = 1.57s

x=0+6.94(1.57) = 10.9m

but the actual answer is 9.78

why am i off?

 

[Delphi51]

I didn't check your calcs - got frustrated when I couldn't find the mass, angle, etc. It is so nice when people write the full question, word for word! But the method looks good, equations correct except for

0= 5.25 + 1/2(-9.8)t

This might be the d = di + .5*a*t² formula.
If so, you are missing the square.

 

[dantra]

That's correct, thank you for your help.

Dan