SOLUTION: 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?
(b) When does
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-> SOLUTION: 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?
(b) When does
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Question 777517: Use the formula
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?
(b) When does the ball reach a height of 48 ft?
(c) What is the greatest height reached by the ball?
(d) When does the ball reach the highest point of its path?
(e) When does the ball hit the ground? Found 2 solutions by stanbon, solver91311:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Use the formula
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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a) Solve . You will get two zeros for this equation, one will represent the time the ball reaches 24 ft on the way up, and the other will represent the time the ball reaches 24 ft on the way back down.
b) Solve . This equation has a conjugate pair of complex zeros. What do you suppose that means? Wait to answer this one until after you have answered c) and d).
Do d) before you do c)
d) The -coordinate of the vertex of is given by . So calculate . This is the time of the highest point.
c) Use the result of part d), , to calculate at time .
NOW answer the question: "When does the ball reach a height of 48 ft?"
e) Solve for the non-zero root. (time zero is when you threw the ball in the first place.)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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