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| Question 1087912:  Hello, I am having trouble answering two questions on projectiles using formulas: d=16t^2 where d=distance in feet that an object falls, in t seconds, regardless of weight.  And, if a projectile is launched upward from the ground with an initial velocity of 73.5 m per second, neglecting air resistance, it's height, in meters, above the ground t seconds after projection is given by the formula: d= -4.9t^2+73.5t.
 *After how many seconds will the projectile be 100 m above the ground? I took 100 and divided by 73.5 and got 1.36 seconds?
 *How long will it take for the projectile to return to the ground?  I converted the 100 m to 328.084 ft and used d=16t^2 to get 4.53 seconds?  Could someone please tell me if these are correct?  If not, could you please help me figure this out?  Any help would be most appreciated! Thanks!
 Found 3 solutions by  solver91311, ikleyn, josmiceli:
 Answer by solver91311(24713)
      (Show Source): Answer by ikleyn(52879)
      (Show Source): Answer by josmiceli(19441)
      (Show Source): 
You can put this solution on YOUR website! For the question: " After how many seconds will the projectile be 100 m above the ground? "
 the units are meters and seconds, so all you need is the formula
 
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 The formula is a parabola with a peak, so unless the projectile is at its
 peak at 100 m above ground, there will be 2 solutions, one when the projectile is
 on the way up, and another on the way down.
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  m Plug this value into the formula
 
   
   Use the quadratic formula
 
  
  
  
  
  
  
  
  
  
  sec and, the other solution is:
 
  
  
  sec These are the 2 solutions
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 " How long will it take for the projectile to return to the ground? "
 At ground level,
  
   
  
  This is when the projectile is launched 
  
  
   In 15 sec, the projectile hits the ground.
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 I can use this answer to check my 1st answer.
 The peak will be at 1/2 this time, or
  sec ---------------------------------------------
 My solutions for the height of
  m must be equally spaced on either side of
  sec, so I can say: 
  
  Right on the money!
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 Here is a plot of this problem:
 
  
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