Question

A stone is thrown vertically into the air from a tower 110 ft. high at the same time that a second stone is thrown upward from the ground. The initial velocity of the first stone is 60 ft/s and that of the second stone 85 ft/s. When and where will the stones be at the same height from the ground?

Answer 1

The stones will be at the same height of 62.59 feets 4.4 seconds later.

Step-by-step explanation:

We will be using the following kinematic equation for 1D movement:

`y(t) \ = \ y_0 \ + \  v_0 \ t  \ + \ (1)/(2) \ a \ t^2`.

For the first stone we got:

`y_a(t) \ = \ 110 \ ft \ + \  60 (ft)/(s) \ t  \ - \ (1)/(2) \ g \ t^2`,

of course, taking the gravitational acceleration

`a \ = \ - \ g \ = \ - \ 32.17 \ (ft)/(s^2)`.

For the second stone we got:

`y_b(t) \ = \ 0 \ ft \ + \ 85 (ft)/(s) \ t  \ - \ (1)/(2) \ g \ t^2`.

`y_b(t) \ = \ 85 (ft)/(s) \ t  \ - \ (1)/(2) \ g \ t^2`.

Now, we want to find the time t' at which both stones will be at the same height, this is

`y_a(t') = y_b(t')`.

So

`\ 110 \ ft \ + \  60 (ft)/(s) \ t'  \ - \ (1)/(2) \ g \ t'^2 = \  85 (ft)/(s) \ t'  \ - \ (1)/(2) \ g \ t'^2`.

Working a little the equation

`\ 110 \ ft \ + \  60 (ft)/(s) \ t'  - \ 85 (ft)/(s) \ t'  \ - \ (1)/(2) \ g \ t'^2 + (1)/(2) \ g \ t'^2 = 0`

`\ 110 \ ft \ - \  ( 25  (ft)/(s) ) \ t' = 0`

`\ 110 \ ft \ = \  ( 25 (ft)/(s) ) \ t'`

`\ (110 \ ft)/( 25 (ft)/(s) ) \ = \ t'`

`4.4 s = \ t'`

Now, to find where the stones will be, we can just replace t for t' in any of the two formulas for the stones:

`y_a(4.4 \ s) \ = \ 110 \ ft \ + \  60 (ft)/(s) \ 4.4 \ s  \ - \ (1)/(2) \ 32.17 \ (ft)/(s^2) \ (4.4 \ s)^2`

`y_a(4.4 \ s) \ = 62.59 ft`