SOLUTION: A ball is thrown upward with an intial velocity of 48foot/second from a height of 640ft. H=-16t^2+48t+640. After how many seconds will the ball hit the ground? (i.e. for what posit
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Question 266443: A ball is thrown upward with an intial velocity of 48foot/second from a height of 640ft. H=-16t^2+48t+640. After how many seconds will the ball hit the ground? (i.e. for what positive value of t is h=0?) Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A ball is thrown upward with an intial velocity of 48foot/second from a height of 640ft. H=-16t^2+48t+640.
After how many seconds will the ball hit the ground? (i.e. for what positive value of t is h=0?)
:
We have
-16t^2 + 48t + 640 = 0
Simplify and change the signs to make it easier to factor, divide by -16
t^2 - 3t - 40 = 0
Easily factors to
(t - 8)(t + 5) = 0
positive solution
t = 8 sec to hit the ground
;
;
Check solution in original equation
-16(8^2) + 48(8) + 640 =
-1024 + 348 + 640 = 0
