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Tossing Your Lunch

  1. Sep 10, 2004 #1
    Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.55m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 37.6m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 6.00s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

    my question: With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? Take free fall acceleration to be g=9.80m/s^2 .

    #2) Where is Henrietta when she catches the bagels? Take free fall acceleration to be g=9.80m/s^2 .

    can someone help me with telling me what formula i need? and can you explain why you picked the formula? i always have trouble picking formulas
  2. jcsd
  3. Sep 10, 2004 #2
    Rather than knowing which formulas to apply when, it is important in physics to learn how to use basic principles and methods of thinking to solve problems. Try to think of the horizontal and vertical components of the path of the bagels separately. Here's what you know: vertically, they travelled a distance of 37.6 m while being accelerated by gravity and with an initial velocity of 0 m/s; horizontally, they maintained a constant speed until they were caught by Henrietta. Try to go from there.
  4. Sep 11, 2004 #3
    x(t) = x(0) + v(0)t + 1/2at^2
    37.6 = 0 + 0 + 1/2(-9.8)t^2

    using that formula i got the time it took for the bagel to drop, which is t=2.77s

    ok i know that Henrietta is currently at t=6.00s. what do i do next? i was trying to figure it out, but i dont get it
  5. Sep 12, 2004 #4
    You know Henrietta's speed and how long she's been walking (6s+2,77s) when the lunch is at ground level. Now you just need to know how fast the the lunch needs to move horizontally to cross that same distance in 2,77s (the time it takes for it to hit the ground [Henrietta] and thus stop moving).
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