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

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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


[itex]x=x_0+vt+\frac{at^2}{2}[/itex]

The Attempt at a Solution


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

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

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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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.