# How to solve a velocity problem as an initial condition problem

## Homework Statement

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 30.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.

## Homework Equations

Acceleration of gravity = 9.81m/s^2

The normal way:
v^2 = u^2 + 2as
v=u+a*t

How I want to do it:
dv/dt = a(t)
ds/dt = v(t)

## The Attempt at a Solution

a(t) = -9.8 m/s^2
v(t) = -9.8t m/s + C(m/s)
17 = -9.8*0 + C, C=17
v(t) = -9.8t(m/s) + 17(m/s)
s(t) = -4.9t^2 + 17t + C
30 = -4.9*0 + 17*0 + C
C = 30
s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

t=5.64s for part B.
For part A, this would make it -9.8*5.64 + 17 = -38.272.

v=(17^2+2*9.81*30)^(1/2)=29.6m/s and 29.62=-17+9.81*t, t=4.76s

I'm certain there's a way to solve this as initial value problems. I'd rather get my calculus on than try to memorize these equations!

s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

Why -30 on the left hand side?

I believe a bad guess. The initial position is y=0m (the level of the roof), the ball ends up at y=-30.

That probably should be a zero.

You need to decide. Either s(0) = 30, which fixes one value for the integration constant. Or s(0) = 0, which fixes some other value. Then you should have s(T) = 0 or S(T) = -30, respectively. But s(0) = 30, which you chose originally, is not compatible with with s(T) = -30 (because that would mean the street is 60 meters below the 30 meter tall building, which is impossible).

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Thank you! That does provide the correct answer in either case (t=4.76, then that can be substituted into v(t) = -29.6m/s!)

I'm glad to know it was something simple.