1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Best time to throw a ball

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    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)

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

    3. 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?
     
    Last edited: Nov 2, 2015
  2. jcsd
  3. Nov 2, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  4. Nov 2, 2015 #3
    Sorry I am new here, what does SUVAT means?
     
  5. Nov 2, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    What is the definition of "best" for this problem? What criteria are to be maximized or minimized?
     
  6. Nov 2, 2015 #5
    so the ball will reach the highest height, I have edit the question and added it
     
  7. Nov 2, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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
     
  8. Nov 2, 2015 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

  9. Nov 3, 2015 #8
    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}##
     
  10. Nov 3, 2015 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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 cancelled out), but you effectively derived that by combining two others.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Best time to throw a ball
  1. Throwing a ball (Replies: 9)

  2. Throwing Balls (Replies: 3)

Loading...