Posted by **Jenny** on Tuesday, August 22, 2006 at 10:22pm.

A mass is dropped from a height of 30m, and at the same time from ground level (and directly beneath the dropped mass), and another mass is thrown straight up with a speed of 15 m/s.

a) How far apart are the two masses after 0.5 s?

b) Will the massses ever meet or pass each other? And if so will the meeting place take place while both masses are falling or when one is falling and the other is rising?

a) solve the distance equation for each particle. I suggest this form..

y=Yo + Vo time + 1/2 a time^2

Use Positive quantities for up, and negative for down.

I will be happy to critique your thinking.

y=Yo + Vo time + 1/2 a time^2

0 = 30+0t =1/2 (-9.8)t^2

0=30 + 1/2 (-9.8)t^2

t^2 = -30/(-9.8* 2)

t=1.237 seconds

y=Yo + Vo time + 1/2 a time^2

0 =0 + (15m/s)t +1/2 (-9.8)t^2

t=3.06122 seconds

Did I do this correctly?

A mass is dropped from a height of 30m, and at the same time from ground level (and directly beneath the dropped mass), and another mass is thrown straight up with a speed of 15 m/s.

a) How far apart are the two masses after 0.5 s?

b) Will the massses ever meet or pass each other? And if so will the meeting place take place while both masses are falling or when one is falling and the other is rising?

a)

y=Yo + Vo time + 1/2 a time^2

0 = 30+0t =1/2 (-9.8)t^2

0=30 + 1/2 (-9.8)t^2

t^2 = -30/(-9.8* 2)

t=1.237 seconds

y=Yo + Vo time + 1/2 a time^2

0 =0 + (15m/s)t +1/2 (-9.8)t^2

t=3.06122 seconds

Did I do this correctly?

Then how will i figure out how far apart the two masses are after .5 seconds? Also How can I figure out part b?

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