SOLUTION: Please, please help me, I have been working on this problem for 1 1/2 hours and cannot solve it. God Bless! Lin An archery enthusiast was testing his new bow. He shot an arrow

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Question 95519: Please, please help me, I have been working on this problem for 1 1/2 hours and cannot solve it. God Bless! Lin
An archery enthusiast was testing his new bow. He shot an arrow straight up from the top of a building 48 feet high with an initial velocity of 32 feet per second. The height of the arrow (h measured in feet) after t seconds in flight is described by the equation
h = 48 + 32t - 16t^2.
Determine the number of seconds after the arrow is shot that it will strike the ground.
t = ??? seconds

Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
When it strikes the ground the height = zero.
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48 + 32t - 16t^2 = 0
16t^2-32t-48 = 0
Divide thru by 16 to get:
t^2-2t-3=0
Factor to get:
(t-3)(t+1)
Positive answer:
t = 3 seconds
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The arrow will be on the ground 3 seconds after it was shot up.
Cheers,
Stan H.

Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
When the arrow hits the ground its height will be 0.
-16t^2+32t+48=0
multiply -1 times both sides. 16t^2-32t-48
Factor 16(t^2-2-3)=16(t-3)(t+1)
t=3 t=-1
3 sec (Ans.)
EdJones