a soccer ball bounces off a players head, straight into the air, from a height of 6ft.........

Question

a soccer ball bounces off a players head, straight into the air, from a height of 6ft, with initial velocity of 40 ft/sec. how long(to the nearest tenth of a second) does it take for the ball to hit the ground. use the function h(t)=-16t^2+40t+6

 

 

 

with steps please 

Answer 1

h(t) = -16t² + 40t + 6

 

This is a quadratic equation.  We want to find when the height is zero.

 

-16t² + 40t + 6 = 0    solve for t

We can factor a 2 out of each term

2(-8t²+20t+3) = 0

Divide each side by 2

-8t²+20t+3 = 0

 

Now we can use the quadratic equation with smaller numbers:  t=-b/2a ±[√(b²-4ac)]/2a

with a = -8, b = 20, c = 3

 

t =  -20/2(-8) ± [√(20² -4(-8)(3))]/2(-8) 

 = -20/(-16) ±[√(400 + 96)]/(-16)

 = 5/4 ± √496/(-16)

 = 5/4 ± √((16)(31))/(-16)

 = 5/4 ± 4√31/(-16)

 = [5 ± -√31]/4

 = (5 + √31)/4   or   (5 - √31)/4

 = 2.6419 sec   or  -0.14194 secs 

 

Rounded to the nearest tenth of a second and disregard the negative answer.

The soccer ball will hit the ground approximately 2.6 seconds after the bounce.

 

When we check the rounded answer in the quadratic we end up with h = 1.84 instead of h = 0

However; checking the answer as given above with t = 2.6419secs we get h = 0.00183 so our answer is correct.