Wednesday

March 4, 2015

March 4, 2015

Posted by **Henry** on Thursday, August 24, 2006 at 9:53pm.

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?

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?

for a: I dont understand how you set y to zero. Y is what you are solving for at t=.5 sec. So solve for y at that time for each ball, and see where they are. The difference between those y is how far apart they are.

for b: set each y equation equal to the other, and solve for time.

30 +1/2 (-9.8)t^2 = (15)t +1/2(-9.8)t^2

a) I got 8.725 for the ball dropped. I got 6.275 for the ball thrown up .

So 2.45 would be how far apart they are?

b)30 +1/2 (-9.8)t^2 = (15)t +1/2(-9.8)t^2

30 + -4.9t^2 = 15t + -4.9t^2

30=15t +o

t= 2s

Are these correct?

a/ I didnt check the math, but yes, the difference in the y's is how far apart they are. Isnt the unit meters?

b. at t=2 they meet.

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