Kinematics - projectile motion - time to maximum height?

Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving projectile motion, specifically focusing on the time it takes for a water drop to reach its maximum height when discharged vertically from a fire hose at a speed of 10 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the logic of breaking the problem into steps, questioning whether finding maximum height is necessary for calculating time. Some attempt to apply kinematic equations to derive time, while others express confusion about the implications of the results and the nature of projectile motion.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the validity of their reasoning. Some guidance has been offered regarding significant figures and the interpretation of results, but there is no explicit consensus on the correct approach or understanding of the concepts involved.

Contextual Notes

Participants note that the problem involves multiple choice answers and significant digits, which may influence their calculations. There is also confusion regarding the definition of a parabola in relation to projectile motion and the conditions under which maximum height is reached.

kenji1992
Messages
22
Reaction score
0
Kinematics -- projectile motion -- time to maximum height?

Homework Statement


The nozzle of a fire hose discharges water at a speed of 10 m/s. The nozzle is aimed straight up. How long does it take for a water drop to reach its maximum height?
Start with translating the question:
vi= 10 m/s
vf= 0 m/s
t=? s
a= -9.8 m/s^2
delta−x= ?

Homework Equations





The Attempt at a Solution



OK, so I'm assuming that solving this takes 2 steps. One entails finding the maximum height, then using a kinematic formula to find time. Is that logic sound?
 
Physics news on Phys.org
What are the relevant equations?

ehild
 
I tried this:
Vfy=Viy+Ay*t
0 m/s=10 m/s-9.8 m/s^2*t
-10 m/s=-9.8 m/s^2 *t
t=10 m/s / -9.8 m/s^2
t=1.02 s
 
It is correct. You see, the maximum height was not needed. ehild
 
1.02 s is not an option for this multiple choice problem, the closest is 1.0 s. Is that likely the answer then?

Edit: The question asks how long it takes to reach maximum height, so because it is a parabola, it stands to reason that 1.02s/2=0.51s is how long it takes to reach the maximum height. Right?
 
The data in the problem are given with two significant digits. Round of your result to t=1.0 s. .

What is parabola?

When does a body projected straight upward reach maximum height? What is its velocity at the apex?
What is the maximum height?

What would be the velocity at t=0.5 s? Would the water drop move upward or downward then? Does it reach the maximum height at t=0.5 s?

ehild
 
I don't know. I remember someone in class saying that projectile motion was a parabola...

t=1.0 s means what? The time it takes for the motion to finish? or the time at which the water reaches its maximum height?
 
The motion is not parabola. A graph of a function can be parabola.

You applied the equation Vfy=Viy+Ay*t. What do the letters mean? You substituted Vfy=0. If you throw up a pebble, will it reach the ground with zero velocity? Where is the velocity zero?
Just throw up something and see...

ehild
 
kenji1992 said:
I don't know. I remember someone in class saying that projectile motion was a parabola...

The problem statement says "The nozzle is aimed straight up".
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
40
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
16
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K