Answer to Question #318536 in Classical Mechanics for renzo08

Question

1.     A stone is thrown vertically upward from ground level. Six seconds later, a second stone is also thrown vertically upward from the same point with a velocity of 40 m/sec. If the two stones meet 30 meters above the ground with the first stone already going down, with what velocity was the first stone thrown?

Answer 1

$v_2=gt-v_1+\sqrt{2gh}=20~\frac ms.$