A ball is thrown from a height of 110 feet with an initial downward velocity of 6/fts .

Question

The ball's height h (in feet) after t seconds is given by the following.

How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth.

h=110-6t-16t^2

 

Answer 1

Rearrange equation.

h=-16t^2-6t+110

 

h = 0 at the ground.

 

divide both sides of the equation by (-2) to yield:

 

0 = 8t^2+3t - 55

 

and use quadratic equation h=-b±√(b² - 4ac)

                                                    2a

where a = 8, b= 3, c=-55

 

Substitute:  [-3 ± √(9 - 4*8*(-55))] / (16)

 

= [6 ± √(36 + 1760)] / (16)

 

= [6 ± √(1796)] /16

 

= (6 ± 42.3792)/16.  Time can only be positive for this problem...so discard negative answer.

 

= 48.3792/16 = 3.02 seconds