Physics challenger question help!?

C

carbonGEISHA

New member
Hi there
these questions are going to be on my exam tomorrow, and I'd really like some help with them (I've been trying for over a half hour and can't seem to quite get them right)
help would be greatly appreciated!
here are the questions:
-A stone is dropped from the top of a 95.0 m building. A second stone is dropped 0.75 seconds later. How far from the ground is the second stone as the first stone hits the ground?

-A student throws a stone vertically upward from the base of a 51 m building. At the same time, a second student on the top of the building drops a stone. If the two stones collide 12 m from the ground, what was the initial velocity of the thrown stone?
Thank you so much guys!
 
oops, I misread the problem, so I'll have to retype.
calculate how long it takes for 1st to hit the ground.
s = (gt^2)/2, where s=95m
solving for t, t = sqrt(2s/g), plug in the values and evaluate
this is the total time of the 1st stone's fall, if you subtract the 0.75 seconds from this time, that will give you the duration of time that the 2nd stone has been falling until the 1st stone hits the ground. so use the same equation, but find s for this value of time. but this is the distance the stone fell. subtract it from 95m to find how far from the ground it is. thus the final answer is:
x = 95m - (9.81m/sec^2[sqrt(2 * 95m/9.81m/sec^2) - 0.75 sec]^2)/2 = 29.6m

notice how I found intermediate answers only algebraically, and actually waited until my final equation to plug in the numbers.

now in the second problem, we have to think very carefully so we don't get confused. let's write an equation for 1st stone:
 
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