Kinematics projectile motion problem

Click For Summary

Homework Help Overview

The problem involves a ball thrown from a height of 490 meters with an initial velocity of 200 m/s. The main question is to determine how far from the base of the building the ball will land, considering the effects of projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for two equations to solve the problem, focusing on time to impact and horizontal distance traveled. There is confusion regarding the application of the kinematic equation and the role of time in the calculations. Some participants question the direction of the initial velocity and its implications for the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have suggested that the initial velocity is horizontal, while others have pointed out the ambiguity in the problem statement regarding the direction of the throw. No consensus has been reached yet.

Contextual Notes

There is uncertainty regarding the initial direction of the ball's throw, which significantly affects the analysis. Participants note that the problem lacks clarity on whether the throw is vertical, horizontal, or at an angle.

tolu
Messages
2
Reaction score
0
1. The problem statement, all variables and given/known
A ball is thrown from atop a 490m building with an initial velocity of 200m/s how far from the bottom of the building will it land


Homework Equations


d=vit+1/2at^2


The Attempt at a Solution


i have tried to put this in a equation and it was 490=-4.9 and it was 100 time is always multiplied by two to get x and it was 200 frm there i am confused
 
Physics news on Phys.org
For one thing, this problem requires TWO equations to solve. You need to find how long it takes for the ball to hit the ground, then find out how far it managed to get in that time.
 
tolu said:
1. The problem statement, all variables and given/known
A ball is thrown from atop a 490m building with an initial velocity of 200m/s how far from the bottom of the building will it land

Homework Equations


d=vit+1/2at^2

The Attempt at a Solution


i have tried to put this in a equation and it was 490=-4.9 and it was 100 time is always multiplied by two to get x and it was 200 frm there i am confused
Well, that equation does not give 490=-4.9. You have dropped t from the equation for no reason. Also, the vit term has disappeared.

MaxL said:
For one thing, this problem requires TWO equations to solve. You need to find how long it takes for the ball to hit the ground, then find out how far it managed to get in that time.
We know that the ball travels 490 m, the height of the building.
 
I should have been more specific! The second step would be to find out how far in the horizontal direction the ball manages to get in that time.
 
Now that I look more carefully, it's actually an ambiguous question. The problem doesn't say the direction in which the ball is thrown--is it straight up? Straight down? Straight away from the building? At a 45 degree angle? The answer definitely depends on the direction the ball is thrown.

If I had to guess, I would say the throw was directly horizontal, since that's a much simpler problem and it seems more in line with where you seem to be in your physics education. Still, you should probably ask your prof. or teacher. Ambiguous questions are not OK!
 
Oops, I was interpreting the question as strictly vertical, and thought it asked how long it takes the ball to drop. My bad.

It does appear that the 200 m/s initial velocity is in the horizontal direction. With no angle given, it would be either vertical or horizontal. But vertical would mean the ball simply lands at the base of the building; it would have to be horizontal for the problem to be meaningful.

So ... initial velocity is 200 m/s in the horizontal direction. What then is the vertical component of the initial velocity?
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
Replies
19
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 18 ·
Replies
18
Views
5K
Replies
8
Views
2K
Replies
7
Views
3K