A ball is dropped from the top of a 45.0 m high cliff.

At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 27.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?

+vel For stone: 45 m = Out Visionet + zat? SBall = ? 45 m = 27m/s (t)+ =(9.81)t2 45m a Ball = - 9.81 mis2 4. 905 t 2 + 276 - 45= 0 4 collide ( t - 1 . 3 403 ) ( t + 6. 8449 ) = 0 H= 1 -D Sstone = 27 mis astone = 9.81 mis ? t = 1 3403s or t= -6. 84495 (rejected) 2 1. 345 For ball