What is the height of a ball after being in the air for T/4 seconds?

  • Thread starter Thread starter robax25
  • Start date Start date
  • Tags Tags
    Ball
Click For Summary

Homework Help Overview

The problem involves a ball tossed directly upward, with a total time in the air denoted as T and a maximum height represented by H. The question seeks to determine the height of the ball after T/4 seconds, while neglecting air resistance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between time and height, with one suggesting that the height is 1/4 of H after T/4 seconds. Others question the timing of reaching maximum height and explore the implications of the equations provided.

Discussion Status

Participants are actively engaging with the problem, with some offering equations and others seeking clarification on the relationships between variables. There is a focus on understanding the motion of the ball and how to apply the equations correctly, though no consensus has been reached on the exact height at T/4 seconds.

Contextual Notes

There is mention of potential confusion regarding the use of different notations for acceleration and the need to clarify the initial velocity. Participants are also considering the effects of the ball's motion in both ascending and descending phases.

robax25
Messages
238
Reaction score
3

Homework Statement



A child tosses a ball directly upward. Its total time in the air is T. Its maximum height is H. What is its height after it has been in the air a time T/4? Neglect air resistance.

Homework Equations


y=volt +0.5 at²

The Attempt at a Solution


It is 1/4 H but I don't get exact answer.
 
Last edited by a moderator:
Physics news on Phys.org
robax25 said:

Homework Statement



A child tosses a ball directly upward. Its total time in the air is T. Its maximum height is H. What is its height after it has been in the air a time T/4? Neglect air resistance.

Homework Equations


y=volt +0.5 at²

The Attempt at a Solution


It is 1/4 H but I don't get exact answer.

What answer do you get? Please show.

At what time do you think the ball reaches the maximal height H?
 
Last edited by a moderator:
t/2 second
 
robax25 said:
t/2 second

What value are you able to obtain with that information and the equation you posted in section 2 of the template in your first post?
 
y=v(t/2)-0,5a(t/2)² when the ball is at maximum height and vo is equal to gt/2
 
robax25 said:
y=v(t/2)-0,5a(t/2)² when the ball is at maximum height and vo is equal to gt/2

So, you can find ##t## from the first equation (using ##y = H##, a given input in the problem); then, after finding ##t## you can figure out ##v_0##. Take it from there.

By the way: either use ##g## or ##a##, but not both notations in the same problem, given that they are both supposed to be the same quantity.
 
robax25 said:
y=v(t/2)-0,5a(t/2)² when the ball is at maximum height and vo is equal to gt/2

Stay with the definition that ##v_0## is the velocity of the ball when it leaves the child's hand, which is what you want to calculate to find the answer of the original problem (##v_0=f\left(g, T, H\right)##).
 
There is no need to find the initial velocity.
robax25 said:
t/2 second
What is t? Why did you add "seconds"?
I guess you mean T/2 - just half of the total air time. That is correct.

Consider the second half of the motion for a moment: How can you describe how far it fell down from the highest point as a function if time?
It drops down by H over a time of T/2. How much does it drop down in half of that time, in T/4? While that is not the final answer to the problem you can transfer that result to get the right answer.
 

Similar threads

Replies
34
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
Replies
12
Views
1K
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
9K
Replies
11
Views
2K