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Homework Help: A ball is thrown upwards . .

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Standing on the ground, you throw a baseball straight upward, releasing the ball at a height of 2.00 meters. The ball travels straight up and then falls(ignoring air resistance). If the ball was in the air for a total of 3.92s, find the maximum height the ball reaches and the speed at which it was thrown.


    2. Relevant equations
    Vf=Vi+at
    (Vf^2)-(Vi^2)=2ax
    x=(Vi*t)+(1/2)(a)(t^2)


    3. The attempt at a solution

    I divided my diagram into 3 velocities:
    velocity initial (Vo):when the ball was released at 2.00 meters
    velocity 2 (V2): velocity at the top of "bell" which is zero
    and the velocity once it hit the ground (Vf)
    Vi=? V2=0 m/s Vf=? i assume Vf is not important

    I am treating time as a net total (3.92) and therefore tnet=3.92=t1+t2
    t1: being the time it take for the ball to reach v2
    t2: being the time it takes for the ball to go from v2(top of the bell with 0m/s) to the ground

    *I treated x the same. Xnet=?=x1+x2
    xi:the distance the projectile traveled until velocity became zero. I am assuming x1=2+?
    x2: the distance the projectile traveled before reaching x=0 meters

    I played around so much with the equations (substituting equations into each other) that I kept getting 2*G as the acceleration. I know thats not right because the ball was thrown upwards and therefore the velocity must be positive.

    The answer ended up being:

    the ball reached 19.8m high
    and 18.7 m/s as the initial velocity.
    I've never had a problem like this that hasn't included an initial velocity.
    o_O
     
  2. jcsd
  3. Sep 13, 2010 #2

    PhanthomJay

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    Gold Member

    You've got the right equations; try your third one to solve for Vi (what is x and what is a and what is t?). Then use your 2nd equation to solve for h.
     
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