A ball is thrown upward from the ground with an initial speed of 35 m/s; at the same instant, another ball is dropped from a building 10 m high. After how long will the balls be at the same height?
2007-08-30 10:25:47 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You have everything you need.
Use y=y0+vt+at^2/2 with proper sign for v and a, |a| =|g|
Find expression of height as a function of time for both balls. Set them equal. Solve for time. Note that the balls might be at the same height when the first ball is on its way up, or it may be when the first ball is on its way down. The easiest way to account for that is to assume that it will be on its way up. If this equation has no solution for time, then it means the first ball was on its way down. Write new equation and solve for time.
2007-08-30 11:32:51 · answer #1 · answered by Snowflake 7 · 0⤊ 0⤋