Simple position-time function question

[tarkovsky]

Homework Statement

A stone falls freely down a well. 1.6 seconds later, another stone is cast 25m/s in a straight line, down the well. The two stones hit the bottom of the well at exactly the same time. How deep is the well?

a_g=9.8 m/s²

Homework Equations

x=x₀+v₀+1/2at²

The Attempt at a Solution

Assuming x₀=0 for both stones and that a/2 is 4.9, my work is as follows

(4.9m/s²)t²=(25m/s)(t-1.6)+(4.9m/s²)(t-1.6)²

Essentially I wanted to find at what point the graphs would intersect. I believe my set up is correct, my problem was solving for the variable t. I was unable to solve for t and was forced to used wolfram to discover the approximate value for t which would represent the total duration for both rocks to strike the ground. I found t≈2.94592s which, when plugged into both equations yielded a result equal at 4 significant figures past the decimal point. x≈42.5243m

Therefore the wells is approximately 42.5243 meters deep.

Is there anyway to solve for x without resorting to wolfram?

 

[tiny-tim]

welcome to pf!

hi tarkovsky! welcome to pf!

(4.9m/s²)t²=(25m/s)(t-1.6)+(4.9m/s²)(t-1.6)²

now expand the brackets, and rewrite in the standard quadratic equation form, ax² + bx + c = 0

 

[PeterO]

Homework Statement

A stone falls freely down a well. 1.6 seconds later, another stone is cast 25m/s in a straight line, down the well. The two stones hit the bottom of the well at exactly the same time. How deep is the well?

a_g=9.8 m/s²

Homework Equations

x=x₀+v₀+1/2at²

The Attempt at a Solution

Assuming x₀=0 for both stones and that a/2 is 4.9, my work is as follows

(4.9m/s²)t²=(25m/s)(t-1.6)+(4.9m/s²)(t-1.6)²

Essentially I wanted to find at what point the graphs would intersect. I believe my set up is correct, my problem was solving for the variable t. I was unable to solve for t and was forced to used wolfram to discover the approximate value for t which would represent the total duration for both rocks to strike the ground. I found t≈2.94592s which, when plugged into both equations yielded a result equal at 4 significant figures past the decimal point. x≈42.5243m

Therefore the wells is approximately 42.5243 meters deep.

Is there anyway to solve for x without resorting to wolfram?

What tiny-tim suggested - then consider the following.

The first stone had a 1.6 second "lead"
If we take g = 10, that means it will reach 16 m/s by then, and have fallen 12.8m
The second stone then begins at 25 m/s, and since they are both accelerating under the influence of gravity - will always be traveling 9 m/s faster.
How long will it take for the second stone to make up the 12.8 metres with an "excess" speed of 9 m/s.
That will tell you when the stones hit the bottom, and the easiest stone to analyse then, is the one that has merely fallen.

Of course, the figures will be slightly different to what I mentioned above, as you should probably use g = 9.8