# Thread: Kinematics Question

1. ## 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.

2. 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 "no-time" 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 ...

$\displaystyle y = 100 - \frac{1}{2}g \left(\frac{10}{\sqrt{2g}}\right)^2$

$\displaystyle y = 100 - \frac{g}{2} \cdot \frac{100}{2g}$

$\displaystyle y = 100 - 25 = 75 \, m$