SOLUTION: 4. A rock is thrown off a 75-meter-high cliff into some water. The height of the rock relative to the cliff after t seconds is given by h(t)=−5t2+20t. Show clear solutions thanks
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-> SOLUTION: 4. A rock is thrown off a 75-meter-high cliff into some water. The height of the rock relative to the cliff after t seconds is given by h(t)=−5t2+20t. Show clear solutions thanks
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Question 1172212: 4. A rock is thrown off a 75-meter-high cliff into some water. The height of the rock relative to the cliff after t seconds is given by h(t)=−5t2+20t. Show clear solutions thanksss!
a. Where will the rock be after five seconds?
b. How long before the rock reaches its maximum height?
c. When will the rock hit the water?
You can put this solution on YOUR website! Draw this, and the equation needed from the water is h(t)=-5t^2+20t+75, the height of the cliff.
after 5 seconds, h(5)=-5*25+20(5)+75=50 m above the water.
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maximum height is at t=-b/2a or -20/-10 or 2 seconds
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it will hit the water when the equation equals 0.
changing all the signs, 5t^2-20t-75=0
there is one positive root
5(t^2-4t-15)=0
t=(1/2)(4+/- sqrt (16+60)); sqrt (76)=8.72
=6.36 seconds
