# Kinematics Question

• Apr 18th 2010, 07:12 AM
RogueDemon
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
$d_1 = 0m$
$a = 9.8m/s^2 [down]$
$v_1 = 0m/s$ Is this velocity actually given? Do I only state a given velocity of 0m/s if "rest" or "stationary" is mentioned?
$d_2 = ?$
$v_2 = ?$
$\Delta{t} = ?$
$\Delta{d} = ?$

Ball
$d_1 = 100m[up]$
$a = 9.8m/s^2 [down]$
$v_1 = 0m/s$
$d_2 = ?$
$v_2 = ?$
$\Delta{t} = ?$
$\Delta{d} = ?$

Required:

$d$ ball and stone meet = ?

Analysis:

Solve for time first.

Ball
$\Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$d_2 - d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$

Stone
$d_2 - d_1 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$

Determine the height at which they meet.

Ball
$\Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$

Verify with stone equation

$\Delta{d} = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$

Solve:

Ball
$d_2 - 100m[up] = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$d_2 - 100m[up] = 0.5a(\Delta{t})^2$
$d_2 = 100m[up] + 0.5a(\Delta{t})^2$

Stone
$d_2 - 0m = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$d_2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$

Sub ball equation into stone equation
$100m[up] + 0.5a(\Delta{t})^2 = v_1(\Delta{t}) + 0.5a(\Delta{t})^2$
$100m[up] = v_1(\Delta{t})$

What do I do next? I think I've made an error with the given $v_1$'s, but I'm not sure how.
• Apr 18th 2010, 08:25 AM
skeeter
any time an object is "dropped" , its initial velocity is zero.

ball's position equation as a function of time ...

$y = 100 -\frac{1}{2}gt^2$

stone's position equation as a function of time ...

$y = v_0 t - \frac{1}{2}gt^2$

set the position equations equal since they have the same position ...

$100 -\frac{1}{2}gt^2 = v_0 t - \frac{1}{2}gt^2$

$100 = v_0 t$ , where $v_0$ is the stone's initial velocity upward.

note the "no-time" equation for constant acceleration ...

$v_f^2 = v_0^2 - 2g(\Delta y)$

at the top of its trajectory, the stone has zero velocity ...

$0 = v_0^2 - 2g(100)$

$v_0 = \sqrt{200g}$

sub in for $v_0$ in the equation $100 = v_0 t$ ...

$100 = \sqrt{200g} \cdot t$

$t = \frac{10}{\sqrt{2g}}$

sub for $t$ in either position equation ...

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

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

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