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?
There are many approaches. Here is one:
Originally Posted by ziggy777
Start the clock from when the the stone is projected. Then:
For the lift: x = (5)(3.5) + 3.5 t = 17.5 + 3.5t .... (1)
For the stone: x = -14t + 4.9t^2 .... (2)
Solve equations (1) and (2) simultaneously for x.
Note: I got equation (2) as follows:
Take downwards as the positive direction. Then:
a = 9.8 m/s^2
u = -14 m/s
t = t
x = ?
For checking purposes:
Originally Posted by mr fantastic
hmmmm..... yes that really solves the problem... many thanks(Hi)