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Homework Help: Easy Vertical Motion Question

  1. Oct 7, 2003 #1
    "Easy" Vertical Motion Question

    This question was on my test today, but it didnt seem right to me..
    Q) An object is thrown vertically at 18m/s from a window and hits the ground 1.6s later. What is the height of the window above the ground?

    Is this even possible?
  2. jcsd
  3. Oct 7, 2003 #2


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    Vertically up or down?

    Use s = ut + 0.5 a*t^2

    You got the t, you got the u, you got the a.

    Seems pretty possible... :wink:
  4. Oct 7, 2003 #3
    Vertically up.
    But how can the entire trip only take 1.6s if the object is shot upwards at 18m/s? Wouldn't it take at least that long to slow the ball down so it could work on its down trip?
  5. Oct 8, 2003 #4


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    Science Advisor

    Were you specifically told "up"? If an object is thrown upwards at 18 m/s, then it will take 18/g= 18/9.8= 1.8 seconds for the object to reach its highest point. Obviously it can't hit the ground before that.

    On the other hand, if the object is thrown vertically downwards at 18 m/s, the distance traveled at time t is x= 18t+ (g/2)t2.
    With t= 1.6 seconds, g= 9.8 m/s2, that gives
    x= 18(1.6)+ 4.9(1.6)2= 41.3 meters.

    Another possibility is that the window is below ground!
    Throwing an object up at 18 m/s gives a height of x= 18t- (g/2)t2. With t= 1.6 and g= 9.8, that gives x= 18(1.6)- (4.9)(1.6)2= 16.3 meters. The ground is 16.3 meters above the window!
  6. Oct 8, 2003 #5
    Thanks for the help everyone!
    It was up and I told my teacher and he took the question off the test........
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