Question 188763This question is from textbook An Applied Approach
: An object is thrown downward, with an initial speed of 16ft/s, from the top of a building 480 ft. high. How many seconds later will the object hit the ground? Use the equation d = vt + 16t2 , where d is the distance in feet, v is the initial speed, and t is the time in seconds.
This question is from textbook An Applied Approach
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An object is thrown downward, with an initial speed of 16ft/s, from the top of a building 480 ft. high. How many seconds later will the object hit the ground? Use the equation d = vt + 16t2 , where d is the distance in feet, v is the initial speed, and t is the time in seconds.
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d(t) = vt + 16t^2
16t + 16t^2 = 480
t + t^2 = 3
t^2 + t -3 = 0
t = [-1 +- sqrt(1-4*-3)]/2
t = [-1 +- sqrt(13)]/2
Positive solution:
t = [2.6056]/2
t = 1.3028 seconds
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Cheers,
Stan H.
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