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Range Question

Discussion in 'MCAT Study Question Q&A' started by Addallat, Sep 24, 2014.

  1. Addallat

    Addallat 7+ Year Member

    Jun 2, 2010
    If a deer is running at 10 m/s, and can maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizontal distance of 20 meters?

    A. 5 m
    B. 10 m
    C. 20 m
    D. 45 m

    Okay, so I've been taught that range = horizontal velocity x time of flight

    therefore if the range is 20 meters and horizontal velocity is 10 meters then time of flight is 2 seconds

    So why can't I plug 2 seconds into the free fall equation y= 1/2gt^2???
    My answer comes out to B

    but the right answer is listed as A 5 meters

    Can someone please explain why I can't plug my time of flight of 2 seconds into the free fall equation of y=1/2gt^2?
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  3. jonnythan

    jonnythan Some men play tennis, I erode the human soul 5+ Year Member

    Jan 26, 2012
    I suppose you're calculating for a fall distance of 10 meters.... but you're failing to realize that if you jump up 5 meters, then fall back down 5 meters, you've "fallen" an equivalent of 10 meters.

    In other words, it takes the same time to fall 10 meters from a standstill that it takes to jump to 5 meters and come back down.
  4. Addallat

    Addallat 7+ Year Member

    Jun 2, 2010
    ahhh thank you

    I'm going to try to explain this to myself:

    so in the equation y = (1/2)gt^2 the t^2 is accounting for Time up and Time down i.e total flight time. But I don't want the total flight time I just want to know how long it takes the deer to get to max height, i.e the top of the flight i.e how high the deer hast to jump

    Since total flight time is 2 seconds and time up = time down, I could therefore plug in 1 second into that equation to get me the right answer of 5 meters for how high the deer has to jump.
  5. labqi

    labqi 2+ Year Member

    May 8, 2013
    The way I did it was:
    1. How long does it take for the deer to cover the horizontal distance? 2 secs. So the total time in the air is 2 seconds. Therefore max height would be at 1 second.
    2. y = vot + 1/2at2
    Where vo = 0 because 0 y velocity thus y = .5x10x1

    Sent from my iPhone using Tapatalk

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