Suppose a rock is thrown upward from a bridge into a river below. The height of

Question

Suppose a rock is thrown upward from a bridge into a river below. The height of the rock above the surface of the water (measured in feet) is given by the function where f(t)--16t2 + 44t + 290 and t represents the number of seconds elapsed since the rock was thrown a. How high is the bridge? 290 feet Preview b. How high above the water (in feet) is the rock 0.5 seconds after it was thrown? « feet Preview c. After how many seconds does the ball hit the water? « seconds Preview d. After how many seconds does the rock reach its maximum height above the water? « seconds Preview e. What is the maximum height of the rock above the water? feet Preview

Explanation / Answer

f = -16t^2 + 44t + 290

b)
t = 0.5 :
f = -16(0.5)^2 + 44(0.5) + 290
f = 308 --> ANS

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c)
-16t^2 + 44t + 290 = 0

Solving by quadratic formula :
We get t = 5.8489

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d)
MAx ht :
We find the vertex...
a = -16 ,b =44

t-value of the vertex = -b/(2a)

= -44 / (2*-16)

= 44/32

1.375

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e)
f = -16t^2 + 44t + 290

f = -16(1.375)^2 + 44(1.375) + 290

f = 320.25 ft

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next problem :

f = -16t^2 + 32t + 155

b)
-16t^2 + 32t + 155 = 0

t = 4.2692

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c)
Again vertex

t-value of vertex = -b/(2a)
= -32/(2*-16)
= - 32 / -32
= 1

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d)
-16t^2 + 32t + 155

Plug in t = 1 :

-16 + 32 + 155

171

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