Question 777517
Use the formula {{{h= -16t^2 + (vo)(t)}}}
A ball is thrown straight upward with an initial speed of vo= 40 ft/s.

(a) When does the ball reach a height of 24 ft?
Solve -16t^2+40t-24 = 0
-2t^2+5t-3 = 0
2t^2-5t+3 = 0
t = [5+-sqrt(1)]/4
t = 3/2 seconds
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(b) When does the ball reach a height of 48 ft?
(c) What is the greatest height reached by the ball?
Vertex occurs when x = -b/(2a) = -40/-32 = 5/4
height = h(10) = -16*(5/4)^2 + 40(5/4) = -25+50 = 25 ft
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(d) When does the ball reach the highest point of its path?
t = 5/4 seconds
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(e) When does the ball hit the ground?
Solve: -16t^2 + 40t = 0
Factor::
8t(-2t+5) = 0
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t = 5/2 = 2.5 seconds
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Cheers,
Stan H.
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