Calculating Projectile Motion for Water Balloon Launch

  • Thread starter Thread starter yaho8888
  • Start date Start date
  • Tags Tags
    Motion
Click For Summary

Homework Help Overview

This problem involves calculating various aspects of projectile motion related to a water balloon launched at a 45-degree angle. The range of the projectile is given as 52 meters, and the time taken to reach that range is 2.9 seconds. Participants are exploring how to find the initial velocity, vertical and horizontal velocities, and the maximum height of the balloon.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of horizontal velocity and its relationship to the initial velocity. There are questions regarding the assumptions made, particularly about air resistance affecting the motion of the water balloon. Some participants suggest using trigonometric relationships to find the initial velocity components.

Discussion Status

The discussion includes various attempts to calculate the horizontal velocity, with one participant successfully determining it as 17.93 m/s. There are ongoing questions about how to derive the vertical component and the maximum height, with some guidance provided on using time and gravitational acceleration to find these values. However, there is no explicit consensus on the final calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The assumption of no air resistance is questioned, particularly in the context of a water balloon launch.

yaho8888
Messages
62
Reaction score
0
This problem is about shooting water Balloons!
If a balloon is fire from an 45 degress angle and it's Range is about 52m and It took about 2.9sec to get to that range. How can I find the initial velocity, the Velocity of vertical, the Velocity of Horziontal, and the Height the Ballon goes?

Show the work please!

THanks!:rolleyes:
 
Physics news on Phys.org
I have found out the horziontal veocity already is 17.93m/s some one please help!
 
The horizontal velocity is [tex]V_0 cos(45)[/tex]
 
You are, I guess. assuming no air resistance. I'm not at all sure that's reasonable for water balloons, but okay. Since there is no horizontal force, the horizontal speed is constant and equal to 52m/2.9s. That's where you got 17.93 m/s. Now, as Sgt said, that' equal to Vcos(45) so you need to solve Vcos(45)= 17.93 for V. Actually, because a 45 degree right triangle is isosceles it should be easy to see that the vertical component of initial velocity is also 17.93 m/s.

Can you get the max. height from that?
 
NO i can't that is the only thing I found out!
 
yaho8888 said:
NO i can't that is the only thing I found out!
Try to equate vertical velocity taking in account initial velocity and acceleration (gravity).
At what time will the velocity be zero? At that instant maximum height is achieved.
 
Thanks I got all of them!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 15 ·
Replies
15
Views
27K
Replies
11
Views
2K
Replies
2
Views
3K
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
18
Views
6K