OK here is a kinematic problem which I simply cannot solve... to be honest it has a really easy solution but I just cannot figure it out. (I know I'm missing something)
A stone is projected vertically upwards with a speed of 14 m/s, from a point O at the top of a mine shaft. Five seconds earlier a mine began to descend the mine shaft from O with a constant speed of 3.5 m/s. Find the depth of the lift (to the nearest metre) at the instant when the stone falls on it. (Neglect air resistance and take g=9.8m/s^2)
I tried to solve this using velocity time graph and god knows what else...
another quick question- when the object is goint vertically upwards g= -9.8 right?? and when it is going down would g=+9.8?