Asked by boggo
Rock A is thrown straight up with an initial speed of 6m/s from the top of a building 52m tall. At the same moment, rock B is thrown straight up from the ground with an initial speed of 25m/s.A) How high above the ground are the rocks when they pass each other? B) If rock B is thrown 2s after rock A, where do they meet?
Answers
Answered by
Damon
Rock A
ha = 52 + 6 t - 4.9 t^2
Rock B
hb = 0 + 25 t - 4.9 t^2
so
52 + 6 t = 25 t
52 = 19 t
====================
ha = 52 + 6 t - 4.9 t^2
hb = 0 + 25 (t-2) - 4.9 (t-2)^2 = -4.9 t^2 + 44.6 t - 69.6
equal when
52 + 6 t - 4.9 t^2 = -4.9 t^2 + 44.6 t - 69.6
38.6 t = 121.6
ha = 52 + 6 t - 4.9 t^2
Rock B
hb = 0 + 25 t - 4.9 t^2
so
52 + 6 t = 25 t
52 = 19 t
====================
ha = 52 + 6 t - 4.9 t^2
hb = 0 + 25 (t-2) - 4.9 (t-2)^2 = -4.9 t^2 + 44.6 t - 69.6
equal when
52 + 6 t - 4.9 t^2 = -4.9 t^2 + 44.6 t - 69.6
38.6 t = 121.6
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.