Building and Technology

2 Responses to “ Physics/math Word Problem? ”

1. 
   gravitat Dec 7th, 2009 at 11:56 am

   The position formula is:
   h(t) = -16t + vt + s
   where v = initial velocity
   and s = initial height
   -16 is the accelleration of gravity (in feet) and stays the same
   In this problem, the initial height is 1250 (top of the building)
   and the initial velocity is positive (“thrown _upward_”) 35
   So this formula becomes:
   h(t) = -16t + 35t + 1250
   The height is the result of the function h(t)
   and time, t, is the input.
   Problems of this nature will do one of three things:
   a) ask for the vertex
   b) give “y” and ask for “x”
   c) give “x” and ask for “y”
   In this case, it asks, “How long will it be before the object hits the ground?”
   “How long” tells us that it is asking for time, t.
   So it has to give height at that time.
   “the object hits the ground” tells us that the height is zero.
   So change the height, h(t), to zero in the function:
   h(t) = -16t + 35t + 1250
   0 = -16t + 35t + 1250
   Since it is set to zero, the quadratic formula can be used at this point.
   0 = -16t + 35t + 1250
   Quadratic formula:
   x = [-b±?(b²-4ac)]/(2a)
   a = -16
   b = 35
   c = 1250
   Plug a, b, and c into the quadratic formula:
   t = [-(35)±?((35)²-4(-16)(1250))]/(2·-16)
   Multiply terms together:
   t = [-35±?(1225+80000)]/(-32)
   t = [-35±?(81225)]/-32
   Get the square root of 81225 using your calculator:
   t = (-35±285)/-32
   Now that you want to add or subtract, split into two parts:
   t = (-35+285)/-32 or t = (-35-285)/-32
   t = (250)/-32 or t = (-320)/-32
   t = -7.8125 or t = 10
   So the time is either -7.8125 or 10.
   Since the negative time would before it was actually thrown, that part isn’t a result that matches the word problem.
   So after ten seconds, it hits the ground.
   —————-
   Check:
   —————-
   Plug 10 in for t:
   h(t) = -16t^2 + 35t + 1250
   h(10) = -16(10)^2 + 35(10) + 1250
   h(10) = -16(100) + 350 + 1250
   h(10) = -1600 + 1600
   h(10) = 0
   Checked!

2. 
   michaele Dec 7th, 2009 at 11:56 am

   0 = -16t^2 + 35t + 1250
   use quadratic formula:
   t = [ -35 +- sqrt(35^2 - 4*(-16)(1250) ] /-32
   solve the above expression and negative time is not admissible.

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