
Kinematics Question
1. A stone is thrown vertically upward from the ground and just reaches the top of a building 100m high. At the same instant that the stone is thrown from the ground a ball is dropped from rest at the top of the building. At what height do the stone and the ball pass one another. (answer: 75m above ground)
Given:
Stone
$\displaystyle d_1 = 0m$
$\displaystyle a = 9.8m/s^2 [down]$
$\displaystyle v_1 = 0m/s$ Is this velocity actually given? Do I only state a given velocity of 0m/s if "rest" or "stationary" is mentioned?
$\displaystyle d_2 = ?$
$\displaystyle v_2 = ?$
$\displaystyle \Delta{t} = ?$
$\displaystyle \Delta{d} = ?$
Ball
$\displaystyle d_1 = 100m[up]$
$\displaystyle a = 9.8m/s^2 [down]$
$\displaystyle v_1 = 0m/s$
$\displaystyle d_2 = ?$
$\displaystyle v_2 = ?$
$\displaystyle \Delta{t} = ?$
$\displaystyle \Delta{d} = ?$
Required:
$\displaystyle d$ ball and stone meet = ?
Analysis:
Solve for time first.
Ball
$\displaystyle \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$\displaystyle d_2  d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
Stone
$\displaystyle d_2  d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
Determine the height at which they meet.
Ball
$\displaystyle \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
Verify with stone equation
$\displaystyle \Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
Solve:
Ball
$\displaystyle d_2  100m[up] = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$\displaystyle d_2  100m[up] = 0.5a(\Delta{t})^2$
$\displaystyle d_2 = 100m[up] + 0.5a(\Delta{t})^2$
Stone
$\displaystyle d_2  0m = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$\displaystyle d_2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
Sub ball equation into stone equation
$\displaystyle 100m[up] + 0.5a(\Delta{t})^2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$\displaystyle 100m[up] = v_1(\Delta{t})$
What do I do next? I think I've made an error with the given $\displaystyle v_1$'s, but I'm not sure how.

any time an object is "dropped" , its initial velocity is zero.
ball's position equation as a function of time ...
$\displaystyle y = 100 \frac{1}{2}gt^2$
stone's position equation as a function of time ...
$\displaystyle y = v_0 t  \frac{1}{2}gt^2$
set the position equations equal since they have the same position ...
$\displaystyle 100 \frac{1}{2}gt^2 = v_0 t  \frac{1}{2}gt^2$
$\displaystyle 100 = v_0 t$ , where $\displaystyle v_0$ is the stone's initial velocity upward.
note the "notime" equation for constant acceleration ...
$\displaystyle v_f^2 = v_0^2  2g(\Delta y)$
at the top of its trajectory, the stone has zero velocity ...
$\displaystyle 0 = v_0^2  2g(100)$
$\displaystyle v_0 = \sqrt{200g}$
sub in for $\displaystyle v_0$ in the equation $\displaystyle 100 = v_0 t$ ...
$\displaystyle 100 = \sqrt{200g} \cdot t$
$\displaystyle t = \frac{10}{\sqrt{2g}}$
sub for $\displaystyle t$ in either position equation ...
$\displaystyle y = 100  \frac{1}{2}g \left(\frac{10}{\sqrt{2g}}\right)^2
$
$\displaystyle y = 100  \frac{g}{2} \cdot \frac{100}{2g}$
$\displaystyle y = 100  25 = 75 \, m$