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.