SOLUTION: The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 128 feet high is s(t) = –16t2 – 32t + 128, where t is the time elapsed that the ball is in t
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-> SOLUTION: The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 128 feet high is s(t) = –16t2 – 32t + 128, where t is the time elapsed that the ball is in t
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Question 1102751: The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 128 feet high is s(t) = –16t2 – 32t + 128, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?
A) -96 ft/sec
B) 256 ft/sec
C) 0 ft/sec
D) 112 ft/sec Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! Instantaneous velocity is defined as ds/dt, or the first derivative of the distance formula
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we want to find t when s(t) = 0
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0 = –16t2 – 32t + 128
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divide both sides of = by -16
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t^2 +2t -8 = 0
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factor the polynomial
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(t+4) * (t-2) = 0
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t = -4 or 2
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we reject the negative value for t
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ds/dt = -32t -32
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now substitute t=2 in ds/dt
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ds/dt = -32(2) -32 = -96
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instantaneous velocity of the ball when it hits the ground is -96 ft/sec
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Answer is A)
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