Block on a frictionless incline

[pcandrepair]

Homework Statement

A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a θ = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

How much time has elapsed between when the block is released and when it hits the floor?

Homework Equations

The Attempt at a Solution

I was required to find the acceleration of the block as it moves down the incline. I found that to be 5.621 m/s^2

I then found the velocity at the point where the block leaves the incline to be 3.13 m/s.

Then I found the distance that the block travels from when it leaves the table to when it hits the ground. This I found to be 1.234 m.

Then I attempted to find the total time of the blocks movement:

Velocity in the y direction:
Vyo = 3.13 Cos(55)
Vyo = -1.7953 m/s

Time for the block from the edge of the table to the floor:
Y = Yo + Vyo - 1/2at^2
0 = 2m - 1.7953t - 1/2(9.8)t^2
t = .48143 sec

Velocity in the x direction:
Vxo = 3.13 Sin(55)
Vxo = 2.56395 m/s


Vf = Vo + at

2.56395 = 0 + 5.621(t)
t = .456 sec

So, the total time would be the two times added together:
.48143 sec + .456 sec = .93743 sec

I submitted this answer and it said it was within 10% of the correct answer...Am I making some sort of rounding error or is there some mistake in my calculations? Thanks for any help!

 

[learningphysics]

"2.56395 = 0 + 5.621(t)
t = .456 sec"

why are you using 2.56395 instead of 3.13 here?

 

[pcandrepair]

That would explain my mistake. I was using the horizontal component of the velocity. When I use 3.13 I get 1.03827 sec, which is correct. Thanks a lot learningphysics!

 

[learningphysics]

That would explain my mistake. I was using the horizontal component of the velocity. When I use 3.13 I get 1.03827 sec, which is correct. Thanks a lot learningphysics!

no prob.