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Homework Help: Projectile motion

  1. Jan 8, 2008 #1
    You throw a ball with launch velocity of [tex]\overrightarrow {v_i}=(3m/s)\widehat{i}+(4m/s)\widehat{j}[/tex] toward a wall, where it hits at height [tex]h_1}[/tex] in time [tex]t_1[/tex] after the launch. Suppose that the launch velocity were, instead, [tex]\overrightarrow {v_i}=(5m/s)\widehat{i}+(4m/s)\widehat{j}[/tex].

    Would the time taken by the ball to reach the wall be greater than, less than, or equal to [tex]t_1[/tex]?

    I thought that the horizontal component played no role in a projectile motion? Answer is less than, I would have chosen equal to. Should I compute the magnitude in order to convince myself other wise (course it would be greater for the 2nd).
     
    Last edited: Jan 8, 2008
  2. jcsd
  3. Jan 8, 2008 #2
    That's the ONLY component that matters here. Horizontal component is faster, gets there faster. When attempting to figure out how long it takes the ball to travel a horizontal distance, you're only interested in how fast it's going horizontally

    Edit: So to make it even clearer, if the problem had the first case being 3 m/s horizontal and 45 sextillion m/s vertically, and the second case is 4 m/s horizontal but 3 nanometers/s vertical, answer is the same(EDIT and by the same I mean the same answer as the original question, which is the second one gets there faster)
     
    Last edited: Jan 8, 2008
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