Solving for Time of Intersection in Vertical Kinematic Problem

[Bryson]

Homework Statement

A stone is dropped of a cliff. A second stone is thrown down the cliff at 10 m/s at 0.5s after the first stone. When do the stones cross paths?

Homework Equations

Yf = vt + at^2/2

The Attempt at a Solution

My logic is the set the final position Yf equal for both stones and solve for time. Keeping in mind for the second stone that t = t + 0.5.

After finding t, simply plug it into the equation to find Yf.

Is this a valid method for solving this problem?

 

[SammyS]

Homework Statement

A stone is dropped of a cliff. A second stone is thrown down the cliff at 10 m/s at 0.5s after the first stone. When do the stones cross paths?

Homework Equations

Yf = vt + at^2/2

The Attempt at a Solution

My logic is the set the final position Yf equal for both stones and solve for time. Keeping in mind for the second stone that t = t + 0.5.

After finding t, simply plug it into the equation to find Yf.

Is this a valid method for solving this problem?

Almost.

Two things:

1. The second stone is in the air for 1/2 second less time than the first stone. You should decide what t stands for precisely.

2. The question asks when the stone cross paths, not where .

 

[paisiello2]

I would define t_total = t1 + t2 and then strictly use t2 in both of your equations:

1) Figure out the value of Yf1 and v1 at t1 = 0.5 for the first object.
2) Yf2 = 0 and v2 is given for the second object.
3) Then set Yf1 = Yf2 both in terms of t2, and solve for t2 as you suggested.
4) Give t_total = 0.5 + t2 as the final answer.