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A ball is thrown up at the edge of a 490 foot cliff.

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A ball is thrown up at the edge of a 490 foot cliff. The ball is thrown up with an initial velocity of 72 feet per second. 
Its height measured in feet is given in terms of time t, measured in seconds by the equation 

h = -16 t^2 + 72 t + 490



a. How high will the ball go and how long does it takes to reach that height?  feet seconds 
b. How long does it take the ball to come back to the ground ?  seconds

asked Oct 8, 2014 in PRECALCULUS by Baruchqa Pupil

1 Answer

+1 vote

a)

For maximum height, v = dh/dt =0

d/dt (-16t² +72t +490 ) = 0

-32t +72 = 0

t = 72/ 32

  = 2.25 sec

Find the maximum height ,

h = -16t² +72t +490

 At t = 2.25 s h is maximum.

h = -16(2.25)² +72(2.25) +490

   = -81 +162 + 490

   = 571 feet

b)

Ball to come back to the ground means h =0

-16t² +72t +490 = 0

Solve this equation

t = {-72 + √[(72)^2 - 4(-16)(490)]}/2(72)

   ={-72 + √[(72)^2 - 4(-16)(490)]}/2(72)

   =(-72 + 191.16 )/144

   = 2.655 s

answered Oct 8, 2014 by bradely Mentor

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