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

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:

$y(t) \ = \ y_0 \ + \ v_0 \ t \ + \ \frac{1}{2} \ a \ t^2$.

For the first stone we got:

$y_a(t) \ = \ 110 \ ft \ + \ 60 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2$,

of course, taking the gravitational acceleration

$a \ = \ - \ g \ = \ - \ 32.17 \ \frac{ft}{s^2}$.

For the second stone we got:

$y_b(t) \ = \ 0 \ ft \ + \ 85 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2$.

$y_b(t) \ = \ 85 \frac{ft}{s} \ t \ - \ \frac{1}{2} \ g \ t^2$.

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

$\ 110 \ ft \ + \ 60 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2 = \ 85 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2$.

Working a little the equation

$\ 110 \ ft \ + \ 60 \frac{ft}{s} \ t' - \ 85 \frac{ft}{s} \ t' \ - \ \frac{1}{2} \ g \ t'^2 + \frac{1}{2} \ g \ t'^2 = 0$

$\ 110 \ ft \ - \ ( 25 \frac{ft}{s} ) \ t' = 0$

$\ 110 \ ft \ = \ ( 25 \frac{ft}{s} ) \ t'$

$\ \frac{110 \ ft}{ 25 \frac{ft}{s} } \ = \ t'$

$4.4 s = \ t'$

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

$y_a(4.4 \ s) \ = \ 110 \ ft \ + \ 60 \frac{ft}{s} \ 4.4 \ s \ - \ \frac{1}{2} \ 32.17 \ \frac{ft}{s^2} \ (4.4 \ s)^2$

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