A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown.What is the bridge's height above the water?feet How many seconds after being thrown does the rock hit the water?secondsHow many seconds after being thrown does the rock reach its maximum height above the water?secondsWhat is the rock's maximum height above the water?

Question
Asked Nov 4, 2019

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown.

  1. What is the bridge's height above the water?

    feet

 

  • How many seconds after being thrown does the rock hit the water?

    seconds

  • How many seconds after being thrown does the rock reach its maximum height above the water?

    seconds

  • What is the rock's maximum height above the water?

 

check_circleExpert Solution
Step 1

It is given that the function f(t) = -16t2+44t+88 determines the height of the rock above the surface of water.

1.The height of the bridge above water is determined.

When 0, the height of the bridge is obtained.
That is,f(
16/2 +441 +88
f(0)=-16(0)+44(0) + 88
Hence the height of the bridge is 88 feet
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When 0, the height of the bridge is obtained. That is,f( 16/2 +441 +88 f(0)=-16(0)+44(0) + 88 Hence the height of the bridge is 88 feet

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Step 2

2.The time taken for the rock thrown to hit the water is computed.

As the rock hits the water then the functionf(t) is zero.
.f()-16441 +88 implies,
-16t244t88 = 0
11 473
t
(by using quadratic formula)
8
= 4.09 seconds
help_outline

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As the rock hits the water then the functionf(t) is zero. .f()-16441 +88 implies, -16t244t88 = 0 11 473 t (by using quadratic formula) 8 = 4.09 seconds

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Step 3
  1. The time taken for the rock to reach the maximum heigh...
The time taken for the rock to reach the maximum height is obtained by
equating the derivative to zero
f(t)=-16/244t+88
f(t)=-321+44
f(t)0 implies,
-32144 0
44
1.375 seconds
32
help_outline

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The time taken for the rock to reach the maximum height is obtained by equating the derivative to zero f(t)=-16/244t+88 f(t)=-321+44 f(t)0 implies, -32144 0 44 1.375 seconds 32

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