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How far did this football go, vector problem!

  1. Oct 12, 2005 #1
    Hello everyone i'm studying for an exam and this problem is a plug in chug probelm but is giving me troubles...
    An athlete throws a shot at an angle of 45 to the hornizontal at an intial speed of 43 ft/sec. It leaves his hand 7ft above the ground.
    (a) where is the shot 2 seconds later
    (b) how high does the shot go?
    (c) where does the shot land?



    here is my work and my confusion.
    http://img291.imageshack.us/img291/3535/lastscan7as.jpg
    THe above is the correct answers and below are the answers i got, i'm confused on why my part (a) the height above the ground, isn't what hte book has, also for part b, it isn't even close to what the book has and im' not sure on how i would approach part c, any help would be great.
     
  2. jcsd
  3. Oct 12, 2005 #2
    I think you're making the simple mistake of missing a term from your equation of vertical motion.

    As for (c), it's a matter of figuring out how long the shot is in the air.
     
  4. Oct 12, 2005 #3
    vertical motion is stated in the book as:
    y = (vosin(angle))t - .5gt^2 and thats what i used
     
  5. Oct 12, 2005 #4
    You've almost got the answer for (a) - you just slipped up on the arithmetic. 7ft - 3.18ft = 3.8 ft to 2 s.f.

    As dylanm says the height the projectile is launched at is usually included in the initial equation so you would have:

    y(t) = h + ut*sin(angle) - 0.5gt^2
     
  6. Oct 12, 2005 #5
    ohhh!! thanks!! whats the ut stand for? so i should have like:
    y(t) = 7+ut*sin(45)-.5(32)(2)^2;
     
  7. Oct 12, 2005 #6
    Sorry should have said, u is the initial speed of the particle and t is the time in seconds. So we have:

    y(2) = 7 + 43*2*sin(45) - 0.5*32*2^2 = 3.8 ft to 2 s.f.
     
  8. Oct 12, 2005 #7
    ahh thank u! :)
     
  9. Oct 12, 2005 #8
    Why does the book have an answer of 21.4 ft for how high it goes when this shows it went 3.8?
     
  10. Oct 12, 2005 #9
    Because the book is asking for the maximum height reached by the projectile over its entire flight and not the height the projectile reached after being in the air for 2 seconds.
     
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