# 1d motion questions

1. Sep 10, 2012

### oreosama

1. The problem statement, all variables and given/known data
a rock is thrown nearly vertical upward from the edge of a tall building of height H. it just misses the edge of the building on the way down and strikes the ground T seconds after being thrown. given H, T

determine the initial velocity of the rock
the max height it reaches above its starting point
the time to reach its max height

2. Relevant equations

v= v_0 + at

x = x_0 + v_0*t + 1/2*a*t^2

v^2 = v_0^2 + 2a(x - x_0)

3. The attempt at a solution

x_0 = H
x = 0
v_0 = ?
v = ?
a = g
t = T

with these as inputs I don't see how I can manipulate algebra to get the things I want. this is a recurring theme with the rest of my homework it seems:

a flowerpot falls off a windowsill and falls past a window below. a person inside the building notices that it takes T seconds to go from the top to the bottom of the window. the window is h meters high. given h, T

determine how far above the window is the windowsill.

v= v_0 + at

x = x_0 + v_0*t + 1/2*a*t^2

v^2 = v_0^2 + 2a(x - x_0)

x_0 = 0
x = q + h (figuring out q is the goal)
v_0 = 0
v = v
a = g
t = T

one again i feel like when i mess with the algebra i end up going in circles where I can't get everything solved within means of the terms given... hurts my head. i think im doing something fundamentally wrong, let me know.

2. Sep 10, 2012

### azizlwl

1. x_0 = H
x = 0
v_0 = ?
v = ?
a = g
t = T
......
x=H
x_0=0

2. x=u2/2a

h=ut+0.5at2

3. Sep 10, 2012

### voko

With $v_0$, the object can reach the maximum height $h$ (relative to the top of the building). That gives you an equation relating $v_0$ and $h$. It takes $T_u$ to lose the initial speed $v_0$ going upward to the maximum height. That gives you an equation relating $v_0$ and $T_u$. It takes $T_d$ to fall from $H + h$ to the ground. That gives you an equation relating $H$, $T_d$ and $h$. And $T_u + T_d = T$. So you have four equations and four unknowns $v_0$, $h$, $T_u$ and $T_d$, which you can solve.