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western50 · Sep 8, 2010, 12:19 AM

The next two questions concern the physical situation shown below. Two balls are are released from a table top of height H = 3m at the same time. The green ball is given an initial velocity v0 = 8 m/s in the downward direction, while the red ball is given an initial velocity v0 = 8 m/s in the upward direction.

https://online-s.physics.uiuc.edu/cg...02/01/hwb2.gif

Compare vbottom(red), the velocity with which the red ball strikes the floor with vbottom(green), the velocity with which the green ball strikes the floor

vbottom(red) < vbottom(green)
vbottom(red) = vbottom(green)
vbottom(red) > vbottom(green)

How long does it take for the red ball to reach the floor?

t = 0.38 s
t = 0.78 s
t = 1.65 s
t = 1.79 s
t = 1.95 s

western50 · Sep 8, 2010, 12:55 AM

Also, I wonder what is the acceleration rate is in this problem?

Unknown008 · Sep 8, 2010, 03:27 AM

If you cannot picture how the balls will move, you can use the formula:

$v^2 = u^2+ 2as$

v is the final velocity, u the initial velocity, a the acceleration due to gravity and s the displacement.

Let's take the downwards motion as positive.

In the green case, $v_r\ ^2 = (u)^2 + 2as$

In the red case, $v_r\ ^2 = (-u)^2 + 2as$

Which of the two will be greater/smaller or neither?

Then, for the second part, use the equation:

$s = ut + \frac12 at^2$

western50 · Sep 8, 2010, 09:22 AM

So the red ball first moves in a parabola way, and then free fall, then how can I be sure about the time/distance the red ball has moved in that parabola way? So, in this case, do I take initial velocity as -8m/s?

western50 · Sep 8, 2010, 09:22 PM

So the red ball first moves in a parabola way, and then free fall, then how can I be sure about the time/distance the red ball has moved in that parabola way? So, in this case, do I take initial velocity as -8m/s?

Unknown008 · Sep 8, 2010, 09:30 PM

If you were to sketch the velocity-time graph for the motion of the red ball, yes, it will have a parabolic shape, and yes, you take initial velocity of the red ball as -8 m/s.

I made a typo in my above post. It's now that I spot it. The velocity for the green ball should read:

$v_g\ ^2 = (u)^2 + 2as$

Here, s means displacement, so the total distance the red ball covered is not important. The displacement is the same because they both started at the edge of the table and ended on the floor.

So, displacement = H m

western50 · Sep 8, 2010, 10:08 PM

so from this equation

$v_g\ ^2 = (u)^2 + 2as$

I am to get what is the value of acceleration, right?