How to solve a velocity problem as an initial condition problem

Click For Summary

Homework Help Overview

The problem involves a rock thrown vertically upward from a building, requiring analysis of its motion under gravity. The context includes initial conditions and the application of kinematic equations, with a focus on understanding the implications of these equations in a calculus framework.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of different equations of motion and the implications of initial conditions on the problem setup. Questions arise regarding the choice of initial position and the resulting integration constants.

Discussion Status

The discussion is active, with participants exploring different interpretations of the initial conditions and their effects on the equations used. Some guidance has been provided regarding the compatibility of chosen values for initial and final positions.

Contextual Notes

There is a noted confusion regarding the initial position of the rock and the final position relative to the building height, leading to questions about the correctness of the setup and calculations.

Wolvenmoon
Messages
11
Reaction score
0

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
http://answers.yahoo.com/question/index?qid=20120204155622AAIs1sA

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.


My answers are wrong. Using the equations listed at Yahoo answers:
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!
 
Physics news on Phys.org
Wolvenmoon said:
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).
 
  • Like
Likes   Reactions: 1 person
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.
 

Similar threads

Replies
2
Views
1K
Replies
11
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 2 ·
Replies
2
Views
872
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K