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With what initial speed does he throw his first ball?

  1. Oct 9, 2004 #1
    1-Balls thrown by a juggler: A juggler performs in a room whose ceiling is at a height h
    above the level of his hands. He throws his first ball vertically upwards so that it just
    reaches the ceiling. At the instant when this happens, he throws his second ball upward
    with the same initial speed.
    (a) With what initial speed does he throw his first ball?
    (b) How much time is required for this ball to reach the ceiling?
    (c) How long a time after the second ball is thrown do the two balls pass each other?
    (d) How far above the juggler’s hands are the balls when they pass each other?
    2-A football kicker can give the ball an initial speed of v . Within what two elevation angles
    must he kick the ball to score a field goal from a point at a distance L to the front of
    goalposts whose horizontal bars is at a height h above the ground?
    (Hint use sin^2θ + cos^2θ = 1 to get a relation between 2 tan^2θ and
    1/cos^2θ substitute
    and then solve the quadratic equation)
  2. jcsd
  3. Oct 9, 2004 #2
    1-(a) mechanical energy theorem implies:
    Em1=Em2 (1:position of the ball when it leaves the hand of the man
    2: when it reaches the ceiling)

    (considering the hand the referential level)

    Em1= epp + ec
    = 0 + 1/2mv^2

    Em2= epp + ec
    = +mgh + 0

    Em1 = Em2

    --> 1/2mv^2 = mgh
    --> 1/2v^2 = gh
    --> v^2 = 2gh (considering g=10 m/s^2)
    --> v^2 = 20h
    --> v = squared root of (20h) m/s

    (b) in this case we have a varied motion for the ball,
    before it reaches the cieling it is a decelerated motion,
    so the equation is:
    z= -1/2gt^2 +vt +v0 ( negative because it's in the opposite way of the mass force
    considering an ascending axiz OZ)

    when -gt + v= 0
    --> t = -v/g
    --> t = -0.1v (s)

    (c) when the second ball leaves the hand of the man
    its equation is z= -1/2gt^2 +vt +v0

    when the first ball leaves the cieling
    its equation is z= 1/2gt^2 +vt

    they will pass each other when

    -1/2gt^2 +v0 = 1/2gt^2
    --> gt^2 = v0
    --> t= squared root of (v0/g) (s)

    (d) because it is the same initial speed
    and considering that the resistance of air is zero.
    we can conclude that the balls are = h/2 m (higher than the juggler)
  4. Oct 9, 2004 #3
    i don't know if it is correct..
    i hope that a mentor can correct it to see if there are mistakes :)
  5. Oct 9, 2004 #4
    for b>
    check the sign of t again ...

    for c>
    is that the total time ? .... ;) u just missed a tiny bit , but rest of the idea is good enough

    for d>
    i haven't checked the calculations but why be unsure?
    can u use the result u got in c?

    -- AI
    p.S > mfk, what have u tried for question 2? do post ur working cuz u can be helped better that way ... the question is straightforward and shouldn't be hard basically ...
  6. Oct 9, 2004 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Before I "correct" anything, I would like to see what mfk_1868 has tried on this problem!
  7. Oct 9, 2004 #6
    i solved these problems but want to see if i solved correctly
  8. Oct 9, 2004 #7
    V/g=t1 V=t1*g 1/2 g*t1^2=h /sqrt {2h/g}=V =>Answer of a)
    t1 = v/g /frac{/sqrt {2h/g}} {g}=t1 =>Answer of b)
    t2 is the time described at c.
    v*t2 - 1/2 g*t2^2 + 1/2 g*t2^2 = h v*t2=h h/v=t2
    t2=/frac{h}{/sqrt {2h/g}} =>Answer of c)

    h - 1/2 g t2^2 =>Answer of d)

    i am not very sure about second problem
  9. Oct 10, 2004 #8
    whatever u have done with question 2 .. be it wrong or right ..
    do post it ..... that way u get ur doubts cleared better ....

    Some initial hints :
    1> forget the hint the question has given for now
    2> Can u come up with a equation that relates the elevation angle (say theta) , L and h?

    -- AI
  10. Oct 10, 2004 #9
    i ll post the second question soon but is my first question true?
  11. Oct 10, 2004 #10
    well i could not understand much of your last post
    but if u have done the same thing A_I_ has done then u are on the right track ...
    i gave a few corrections to his solutions in my last post

    -- AI
  12. Oct 10, 2004 #11
    thanks i ll compare the solutions
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