Answer:
The stones will be at the same height of 62.59 feets 4.4 seconds later.
Explanation:
We will be using the following kinematic equation for 1D movement:
[tex]y(t) \ = \ y_0 \ + \ v_0 \ t \ + \ \frac{1}{2} \ a \ t^2[/tex].
For the first stone we got:
[tex]y_a(t) \ = \ 110 \ ft \ + \ 60 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2[/tex],
of course, taking the gravitational acceleration
[tex]a \ = \ - \ g \ = \ - \ 32.17 \ \frac{ft}{s^2}[/tex].
For the second stone we got:
[tex]y_b(t) \ = \ 0 \ ft \ + \ 85 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2[/tex].
[tex]y_b(t) \ = \ 85 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2[/tex].
Now, we want to find the time t' at which both stones will be at the same height, this is
[tex]y_a(t') = y_b(t')[/tex].
So
[tex] \ 110 \ ft \ + \ 60 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2 = \ 85 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2 [/tex].
Working a little the equation
[tex] \ 110 \ ft \ + \ 60 \frac{ft}{s} \ t' - \ 85 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2 + \frac{1}{2} \ g \ t'^2 = 0[/tex]
[tex] \ 110 \ ft \ - \ ( 25 \frac{ft}{s} ) \ t' = 0[/tex]
[tex] \ 110 \ ft \ = \ ( 25 \frac{ft}{s} ) \ t' [/tex]
[tex] \ \frac{110 \ ft}{ 25 \frac{ft}{s} } \ = \ t' [/tex]
[tex] 4.4 s = \ t' [/tex]
Now, to find where the stones will be, we can just replace t for t' in any of the two formulas for the stones:
[tex]y_a(4.4 \ s) \ = \ 110 \ ft \ + \ 60 \frac{ft}{s} \ 4.4 \ s \ - \ \frac{1}{2} \ 32.17 \ \frac{ft}{s^2} \ (4.4 \ s)^2[/tex]
[tex]y_a(4.4 \ s) \ = 62.59 ft[/tex]
