A rock fell off a cliff and then a second rock fell 3.6s later and the initial velocity is 70.56m/s and the acceleration is 9.8m/s^2. I had to find the time that the rock fell to the ground and I got 5.2s but I think the time is wrong. Then to find the height of the cliff I used a distance formula and got 499m but that answer is wrong if anyone knows what I did wrong that would be very helpful.
2007-10-10 12:46:45 · 3 answers · asked by glance 3 in Science & Mathematics ➔ Engineering
I just ran it out and got 142 m for the height of the cliff and t=5.4, where t is the time of the first rock to hit ground. The second rock will hit at 1.8 seconds after it is thrown.
I think you mixed up your times. When the second rock is thrown at 3.6 seconds, the first rock is has only fallen 63 m and is traveling at 35.3 m/sec. It is not going to take that long for the second rock to overtake.
Good luck
2007-10-10 13:27:51 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 0⤊ 0⤋
If the rock fell off of the cliff and was not propelled or anything i think the initial velocity would have to be 0. Its been a few semesters since i have done this though.
2007-10-10 20:06:26 · answer #2 · answered by ? 1 · 0⤊ 0⤋
jsut dont stand under the rock
2007-10-10 23:10:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋