Calculating Time of Collision for Dropped Stone and Thrown Ball

[clh7871]

A stone is dropped off a cliff of height h. at the same instant a ball is thrown straight up from the base of the cliff with initial velocity (vi). assuming the ball is thrown hard enough, at what time t will stone and ball meet? (neglect air resistance)

I just don't even know how to approach this problem? could someone help me??

 

[jamesrc]

Write the position of each object as a function of time, then set the two expressions equal to each other to solve for time. You should find that they are equal at time t = 0 (the trivial solution to the problem) and some other time that will be your answer. You can find the position of each using the same formula for motion of an object with constant acceleration provided you understand what each term in the formula means:

$$y(t) = y_o + v_{oy}t + \frac 1 2 at^2$$

 

[Diane_]

Write the position of each object as a function of time, then set the two expressions equal to each other to solve for time. You should find that they are equal at time t = 0 (the trivial solution to the problem) and some other time that will be your answer. You can find the position of each using the same formula for motion of an object with constant acceleration provided you understand what each term in the formula means:

$$y(t) = y_o + v_{oy}t + \frac 1 2 at^2$$

Small amendment - the stones don't start from the same place, so the trivial solution t = 0 won't be there. The technique suggested will, of course, still work.

 

[jamesrc]

That's true. Thank you; I must have read the problem too fast.