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Vertical motion Velocity and Acceleration dot products

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is tossed straight up into the air so that it moves in the vertical direction without air resistance. Is there a point where the dot product of the velocity with the acceleration is:
    a. Zero
    b. Negative
    c. Positive
    d. now assume the ball moves in a parabolic path without air resistance. Is there a point where the dot product of the velocity with the acceleration is zero.
    Explain your answers.

    We know that, in vertical motion both ##v_x## and ##a_x = 0##
    We know that in parabolic motion ##a_x = 0##

    2. Relevant equations
    Dot Product ## (v_x , v_y) \cdot (a_x, a_y) ## -> ##(v_x * a_x) + (v_y * a_y)##

    3. The attempt at a solution
    My Assumptions:
    • if y-direction is upwards accleration due to gravity is negative ##a_y=-9.81m/s^2##
    • Velocity when ball is thrown up is positive, and when the ball comes back down the velocity is negative due to change in direction, magnitude is the absolute value so the speed is never negative it is the direction.
    • if y-direction is downwards acceleration due to gravity is positive ##a_y=+9.81m/s^2##

    a. Yes, Assuming y-direction of upwards is positive; ##(0 , v_y) \cdot (0 , a_y)## -> ##(0*0)+(v_y * a_y)## -> ##(v_y * -9.81m/s^2)## -> v_y = 0 at the instant it reaches max height, before the ball travels downwards -> ##(0m/s*-9.81m/s^2)## -> yields value of 0

    b. Yes, Assuming y-direction of upwards is positive; ##(0 , v_y) \cdot (0 , a_y)## -> ##(0*0)+(v_y * a_y)## -> ##(v_y * -9.81m/s^2)## -> v_y = positive when thrown up, -> ##(+m/s*-9.81m/s^2)## -> yields negative value

    c. Yes, Assuming y-direction downward is positive ##(0 , v_y) \cdot (0 , a_y)## -> ##(0*0)+(v_y * a_y)##, -> ##(v_y * 9.81m/s^2)## -> v_y is positive when falling back down, -> ##(+m/s*9.81m/s^2)## -> yields positive value

    d. Yes, assuming y-direction upward is positive, and x-direction to the right is positive. ##(v_x , v_y) \cdot (0, -9.81)## -> no matter the velocity of x the product of x and 0 is still 0 -> ##(v_x*0)+(v_y*-9.81m/s^2)## -> we know that at max height instantaneous velocity of y is zero -> ##(0*0)+(0*a_x)## -> the dot product yield zero

    I'm not sure if my logic/assumptions are correct in regards to c. and d. where the reason why velocity changes is because the direction changes, and when defining y downwards as positive if velocity when the ball is going up if it is positive then, or when it is coming down.
     
  2. jcsd
  3. Feb 26, 2015 #2

    haruspex

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    All good.
     
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