Calculate the time it took the marble to travel from N to P?

[huybinhs]

Homework Statement

A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at L, falls, and hits the ground at P. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from N to P.

Please look at the graph:

http://i995.photobucket.com/albums/af79/huybinhs/plot.png

Homework Equations

p(t) = 0.5 * a * t^2 + v0 * t + p0
where
p(t) is position at time t
a is acceleration
v0 is initial velocity at time 0
p0 is position at time 0

The Attempt at a Solution

p(t) = 0.5 * a * t^2 + v0 * t + p0
where
p(t) is position at time t
a is acceleration
v0 is initial velocity at time 0
p0 is position at time 0

We'll use the fall from L to P to compute v0.

Vertical acceleration due to gravity = -9.8 m/s^2 = -980 cm/s^2
Let t = 0 when the marble is at point L.
horizonatal
px(t) = 100 = 0.5 * 0 * t^2 + v0 * t + 50 = v0 * t + 50
vertical
py(t) = 0 = 0.5 * -980 * t^2 + 0 * t + 90 = -490 * t^2 + 90

Two equations. Two Unknowns. Solve for t and v0.

=> t1 = 0.43; v0= 116.28

To find the time need to travel from point N to L use v0 from above
px(t) = 50 = 0.5 * 0 * t^2 + v0 * t + 10 = v0 * t + 10

=> t2 = 0.34

Add the two times for the total time from L to P.

=> t = t1+t2= 0.774 which came out wrong. I don't know why? Please help and let me know the result!

Thanks so much!

 

[Jebus_Chris]

If you find the time it takes to fall to the ground you can find it's constant horizantal velocity.
.7=\frac{1}{2}gt^2
t=.377

t1 obviously equals t2 since there is no x acceleration and the ball travels the same distance in both parts.

 

[huybinhs]

So you mean final answer is 0.377^2 = 0.142129?

 

[Jebus_Chris]

Um no. Your final answer would be 2*.377 = .754

 

[huybinhs]

Um no. Your final answer would be 2*.377 = .754

I just tried to enter that number (.754). It came out wrong!

Please help!

 

[Jebus_Chris]

Oh, the marble starts at x=10, so the time is slightly less.

 

[huybinhs]

Oh, the marble starts at x=10, so the time is slightly less.

? Could u be more specific and answer please!

Thanks!

 

[Jebus_Chris]

You know the time it takes for the marble to reach the ground, .377. Using that time and the horizontal distance it travels you can find it's x velocity and therefore find out how long it takes to reach point L.

 

[huybinhs]

You know the time it takes for the marble to reach the ground, .377. Using that time and the horizontal distance it travels you can find it's x velocity and therefore find out how long it takes to reach point L.

Now, I have no clue. Please help!

 

[Jebus_Chris]

The time on the first part is
t_1=x_1/v
You can easily find v since you know the time, t2, for the second part and the distance the marble travels.
v = x_2/t_2
Then add up the times.

 

[huybinhs]

The time on the first part is
t_1=x_1/v
You can easily find v since you know the time, t2, for the second part and the distance the marble travels.
v = x_2/t_2
Then add up the times.

Ok. Is this part correct?

p(t) = 0.5 * a * t^2 + v0 * t + p0
where
p(t) is position at time t
a is acceleration
v0 is initial velocity at time 0
p0 is position at time 0

We'll use the fall from L to P to compute v0.

Vertical acceleration due to gravity = -9.8 m/s^2 = -980 cm/s^2
Let t = 0 when the marble is at point L.
horizonatal
px(t) = 100 = 0.5 * 0 * t^2 + v0 * t + 50 = v0 * t + 50
vertical
py(t) = 0 = 0.5 * -980 * t^2 + 0 * t + 90 = -490 * t^2 + 90

Two equations. Two Unknowns. Solve for t and v0.

=> t1 = 0.43; v0= 116.28

 

[huybinhs]

Anyone? Please help!

 

[rl.bhat]

Velocity v at L is 50/0.377 cm/s.
Time to move from N to L is 40cm/v.