Initial velocity, acceleration due to gravity question

Click For Summary

Homework Help Overview

The problem involves a physics lab scenario where a water balloon is tossed upward from a height of 6.2 meters and lands after 2 seconds. Participants are tasked with determining the initial velocity, final velocity, and maximum height of the balloon's trajectory, utilizing equations of motion under the influence of gravity.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find initial velocity and question the setup of the problem. Some express confusion over their calculations, particularly regarding the initial velocity and the time variable.

Discussion Status

The discussion is active, with participants attempting to clarify their understanding of the equations and the problem setup. Some guidance has been offered regarding the correct application of the kinematic equations, and there are varying interpretations of the initial velocity calculations.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the height of release and the time of flight. There is an ongoing examination of the signs and values used in their calculations.

scifi3
Messages
5
Reaction score
0

Homework Statement


In this Lab you will toss water balloons over the edge of a raised sidewalk.
The thrower should hold the balloon over the edge of the sidewalk and then toss it directly upward. As soon as the balloon is released the person on the ground should start the timer. Stop the timer when the balloon hits the ground. Balloon is released at 6.2m above ground and lands in 2 seconds. What is the initial velocity, final velocity and maximum height of travel?

Homework Equations


v=at+Vo
y-y=volt + 1/2at^2


The Attempt at a Solution


Tried to solve for Vo and y by substitution and I keep getting zero!
 
Physics news on Phys.org
"As soon as the balloon is released the person on the ground should start the timer."
That should give you the initial velocity. From there, try solving again.
 
use 0 for time and solve? that just gives me zero for initial velocity again.
 
Your equation should be:

y2 - y1 = v0t + (1/2)at^2

solve for v0. you're given t = 2s
 
Just to make sure I did this right, the initial velocity is 9.8m/s?
 
scifi3 said:
Just to make sure I did this right, the initial velocity is 9.8m/s?

No, that's not what I'm getting. Can you show your calculations? be careful about signs...
 
0-6.2 = v0(2) + (1/2)-9.8(2)^2
v0 = 6.7m/s
 
scifi3 said:
0-6.2 = v0(2) + (1/2)-9.8(2)^2
v0 = 6.7m/s

That's right.
 
great job helping me! thanks
 
  • #10
scifi3 said:
great job helping me! thanks

no prob.
 

Similar threads

Replies
18
Views
2K
Replies
19
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
23
Views
4K
Replies
2
Views
4K
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K