Projectile Motion Equation: Calculating Height from Initial Velocity and Time

Click For Summary

Homework Help Overview

The problem involves projectile motion, specifically calculating the height from which a ball is thrown given its initial velocity and the time it takes to strike the ground. The ball is projected at an angle below the horizontal, which adds complexity to the analysis of its motion.

Discussion Character

  • Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the projectile motion equation, questioning the assignment of signs for the initial velocity and acceleration. There are attempts to clarify how to consistently define positive directions in the context of the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on checking the signs used in calculations. There is an acknowledgment of the need for consistency in defining positive and negative directions, but no consensus has been reached on the specific errors in the original poster's approach.

Contextual Notes

Participants note the importance of correctly assigning signs based on the chosen coordinate system, which is crucial for accurately solving the problem. There is an emphasis on ensuring that the definitions of upward and downward directions are clear and consistent throughout the calculations.

KevinFan
Messages
47
Reaction score
0

Homework Statement


A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90m/s at an angle of 21.9deg below the horizontal. It strikes the ground 2.95s later.
Calculate the height from which the ball was thrown.

Homework Equations

The Attempt at a Solution


y=y0+vy0t- 1/2gt^2
vy0=sin21.9 deg*7.9m/s=2.9466m/s
0=y0+2.9466m/s(2.95s)-(1/2)(9.8m/s^2)(2.95s)^2
y0=33.95m
could anybody tell me where have done wrong?
 
Physics news on Phys.org
Check your signs.
 
  • Like
Likes   Reactions: KevinFan
KevinFan said:

Homework Statement


A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90m/s at an angle of 21.9deg below the horizontal. It strikes the ground 2.95s later.
Calculate the height from which the ball was thrown.

Homework Equations

The Attempt at a Solution


y=y0+vy0t- 1/2gt^2
vy0=sin21.9 deg*7.9m/s=2.9466m/s
0=y0+2.9466m/s(2.95s)-(1/2)(9.8m/s^2)(2.95s)^2
y0=33.95m
could anybody tell me where have done wrong?
You need to be consistent about how you assign the positive directions. If you want positive to be up, what is the sign on y0? If you want positive to be down, what is the sign on the acceleration?
 
  • Like
Likes   Reactions: KevinFan
Bystander said:
Check your signs.
Thank you:)
 

Similar threads

Replies
40
Views
3K
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K