SOLUTION: an object is thrown upward from the top of a 64 foot building with an initial velocity of 48 feet per second. the height h of the object after t seconds is given by the quadratic e

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: an object is thrown upward from the top of a 64 foot building with an initial velocity of 48 feet per second. the height h of the object after t seconds is given by the quadratic e      Log On


   

Question 1139714: an object is thrown upward from the top of a 64 foot building with an initial velocity of 48 feet per second. the height h of the object after t seconds is given by the quadratic equation h= -16t^2+48t+64. When will the object hit the ground?
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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The object will hit the ground exactly and precisely at the time moment "t", when  h(t) = 0,  or, equivalently,


    -16%2At%5E2+%2B+48t+%2B+64 = 0.


Simplify and solve for "t" :


    t%5E2+-+3t+-4 = 0

    (t-4)*(t+3) = 0.


The roots are t= 4  and  t= -3.  Only positive root  t= 4 is meaningful and provides the solution to the problem.


ANSWER.  In 4 seconds.