a stone falls from rest from the top of a high cliff. a second stone is thrown downward frm the same height 2 s later with an initial speed of 30 m/s.if both stones hit the ground simultaneously,how high is the cliff?assume a constant acceleration magnitude g= 9.8 m/s2
2007-09-19 03:54:38 · 4 answers · asked by huwaida b 1 in Science & Mathematics ➔ Physics
dude, do your own physics homework
2007-09-19 04:11:51 · answer #1 · answered by Anonymous · 0⤊ 0⤋
X=x0+v0t+1/2at2 = x0+v0t+1/2at2
9.8*2s = 19.6 = initial velocity of stone 1 when stone 2 is released
1/2 * 9.8 * 4s = 19.6 = initial position of stone 1 when stone 2 is released
19.6 + 19.6*t + 1/2 * 9.8 * t^2 = 0 + 30*t + 1/2 * 9.8 * t^2
19.6 + 19.6*t = 30*t
10.4*t = 19.6
t=1.88s
But add in the 2 seconds from before
total t = 3.88s
now go back to stone 1, x = x0 + v0t + 1/2at^2
x = 1/2 * 9.8 * 3.88^2 = 73.77 m high
2007-09-19 11:09:51 · answer #2 · answered by James H 2 · 0⤊ 0⤋
s = v0(t-2) + (1/2)9.8(t-2)² and
s = (1/2)9.8t² so
(1/2)9.8t² = 30(t-2) + (1/2)9.8(t-2)²
Now do a little algebra and you're done.
Doug
2007-09-19 11:20:36 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
V^2= V0^2 + 2a(x-x0)
X should come out to be 45.91
2007-09-19 11:04:01 · answer #4 · answered by ansaritaha007 2 · 0⤊ 0⤋