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

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The height of a ball after being in the air for T/4 seconds can be calculated using the kinematic equation y = v_0 t + 0.5 a t². Given that the total time in the air is T and the maximum height is H, the ball reaches its maximum height at T/2 seconds. The height at T/4 seconds is determined to be H/4, but further calculations involving initial velocity (v_0) and acceleration (a) are necessary for precise results. It is crucial to maintain consistent notation, using either g or a for acceleration due to gravity.

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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:
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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.
 

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