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Maximum height of launched ball

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is launched at an angle of 28 degrees off a platform. The initial velocity is 53m/s. When dropped straight down, the ball hits the ground after 1.6 seconds.
    What is the maximum height of the ball?


    2. Relevant equations
    Xf = Xi + Vi(t) + (1/2)(a)(t^2)
    Vix = (Vi)(cos@)


    3. The attempt at a solution
    I was able to figure out the height of the platform from which the ball was launched, using the the 1.6s times gravity, to get 12.5m high.

    I figured out that the ball traveled 5.54 seconds by solving for t in the equation Yf = Yi + Viy(t) + (1/2)(g)(t^2) and Viy=(Vi)(sin@). Using this, I found that the ball landed 259.25m away.

    I was thinking I could use the equation of the y/x-graph to find the two X-values when Y = 0, and find Y at the middle X point, to get the highest point of the ball.
    Which equation exactly represents this y/x-graph line? Any other suggestions would be highly appreciated!
     
  2. jcsd
  3. Feb 10, 2009 #2
    You've pretty much got it, though there's no need to graph.

    Initial velocity [itex]v_i_y = v_i sin \theta[/itex]. Due to the ball's path's symmetry, maximum height will be reached exactly halfway, temporally and spatially, between the two points where the ball is 12.5 meters off the ground. So plug in this value for [itex]Y_f[/itex] and [itex]Y_i[/itex] and calculate how much time it takes for the ball to complete this portion of its path. Then just keep in mind that maximum height is reached in half this time.
     
  4. Feb 10, 2009 #3
    Of course!!
    Thank you for your prompt assistance :)
     
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