When is the Best Time to Throw a Ball for Maximum Height?

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Homework Help Overview

The discussion revolves around determining the optimal timing for throwing a ball to achieve maximum height, specifically whether to throw it immediately upon jumping or at the peak of the jump. The subject area includes kinematics and projectile motion, with an emphasis on geometric reasoning without the use of forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore different mathematical approaches to the problem, including the use of SUVAT equations and geometric interpretations. Some question the validity of the original poster's algebraic reasoning, while others seek clarification on terminology and definitions.

Discussion Status

The discussion is ongoing, with participants providing various insights and suggestions for approaching the problem. There is a focus on clarifying definitions and ensuring that the assumptions made in the problem setup are understood. Some participants have offered alternative equations and reasoning pathways, but no consensus has been reached.

Contextual Notes

Participants note the importance of defining what "best" means in the context of the problem, as well as the need to consider the implications of time and gravity in their calculations. There is also mention of potential confusion regarding the application of different SUVAT equations.

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Homework Statement


A kid is throwing a ball to the celling, when will be the best time to throw the ball so the ball will reach the highest height , right after his feet leave the ground or at the peak height of his jump? provide a geometric explanation too (do not uses forces)

Homework Equations


x=x_0+vt+\frac{at^2}{2}

The Attempt at a Solution


Jump and throw immediately: x=x_0+(v_{throw}+v_{jump})t-\frac{gt^2}{2}

Jump and throw in mid air: x=x_0+v_{jump}t-\frac{gt^2}{2}+v_{throw}t-\frac{gt^2}{2}

therefore the "Jump and throw immediately" will provide a higher result, as the deceleration is just once.

As for the geometric explanation I can not find one that does not use mass maybe "Jump and throw" is one parabola and "Jump and throw in mid air" is a half parabola+ another parabola?
 
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I don't find your algebraic justification convincing. You do not know t, and in the second equation the two t's may be different. You need to use a different SUVAT equation.
 
Sorry I am new here, what does SUVAT means?
 
Gbox said:
A kid is throwing a ball to the celling, when will be the best time to throw the ball,...
What is the definition of "best" for this problem? What criteria are to be maximized or minimized?
 
so the ball will reach the highest height, I have edit the question and added it
 
Gbox said:
Sorry I am new here, what does SUVAT means?
The SUVAT equations apply when acceleration is constant. The name comes from the five standard variables in them. Each equation involves a different four of the five variables. To choose an equation for the task, consider which four are of interest.
Usually you will know the values of three and need to evaluate a specific fourth.
Other times, you may have two different motions covered by SUVAT (horizontal and vertical motion of one body; motion in one line by two different bodies) which share some unknown (time, distance), making that unknown "of interest".
https://en.wikipedia.org/wiki/Equations_of_motion#Uniform_acceleration
 
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So I have rethought about it, I am just interested in the max height therefore I should use ##v=u-gt##.
So for we have Jump and throw immediately: ##x_1=x_0+(v_1+v_2)t##
And Jump and throw in mid air: ##x_2=x_0+v_1t_1+v_2t_2##
Because we know that the time depends on the gravity ##v-gt=0## we have ##t=\frac{v}{10}##

and overall Jump and throw immediately: ##x_1=x_0+\frac{(v_1+v_2)^2}{10}##
Jump and throw in mid air: ##x_2=x_0+\frac{v_1^2}{10}+\frac{v_2^2}{10}##
 
Gbox said:
So I have rethought about it, I am just interested in the max height therefore I should use ##v=u-gt##.
So for we have Jump and throw immediately: ##x_1=x_0+(v_1+v_2)t##
And Jump and throw in mid air: ##x_2=x_0+v_1t_1+v_2t_2##
Because we know that the time depends on the gravity ##v-gt=0## we have ##t=\frac{v}{10}##

and overall Jump and throw immediately: ##x_1=x_0+\frac{(v_1+v_2)^2}{10}##
Jump and throw in mid air: ##x_2=x_0+\frac{v_1^2}{10}+\frac{v_2^2}{10}##
Yes, except you lost a factor of 2 somewhere.
Since you do not care about time I would have started with the SUVAT equation v2-u2=2as (basically, energy conservation with the mass canceled out), but you effectively derived that by combining two others.
 

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