Respuesta :

Answer:

2.94 seconds.

Step-by-step explanation:

The ball will hit the ground when the height of the ball is 0 meters.

The equation is...

h = 61 - 6t - 5t^2.

-5t^2 - 6t + 61 = 0

5t^2 + 6t - 61 = 0

We can use the quadratic formula to solve.

[please ignore the A-hat; that is a bug]

[tex]\frac{-b ± \sqrt{b^2 - 4ac} }{2a}[/tex], where a = 5, b = 6, and c = -61.

= [tex]\frac{-6 ± \sqrt{6^2 - 4 * 5 * (-61)} }{2 * 5}[/tex]

= [tex]\frac{-6 ± \sqrt{36 - 20 * (-61)} }{10}[/tex]

= [tex]\frac{-6 ± \sqrt{36 + 1,220)} }{10}[/tex]

= [tex]\frac{-6 ± \sqrt{1,256} }{10}[/tex]

= [tex]\frac{-6 ± 35.44009029}{10}[/tex]

Since time cannot be negative, we will not minus the 35.44009029.

(-6 + 35.44009029) / 10 = 29.44009029 / 10 = 2.944009029

That is approximately 2.94 seconds.

Hope this helps!

Answer:

[tex]\boxed{t = 2.94 \ seconds}[/tex]

Step-by-step explanation:

When the ball hits the ground, h = 0

Putting this in the equation:

=> [tex]0 = 61-6t-5t^2[/tex]

=> [tex]61-6t-5t^2 = 0[/tex]

Taking -1 as common

=> [tex]-1(5t^2+6t-61) = 0[/tex]

Dividing both sides by -1

=> [tex]5t^2+6t-61 = 0[/tex]

Using Quadratic Equation:

=>t =  [tex]\frac{-b+ / - \sqrt{b^2-4ac} }{2a}[/tex]

=> [tex]\frac{-6 +/- \sqrt{6^2-4(5)(-61)} }{2(5)}\\\frac{-6 +/- \sqrt{36+1220} }{10}[/tex]

=> t = [tex]\frac{-6 +/- 35.44}{10}[/tex]

Either:

t = [tex]\frac{-6+35.44}{10}[/tex]       OR     t = [tex]\frac{-6-35.44}{10}[/tex]

t = 29.44/10      OR     t = -41.44/10

t = 2.94            OR      t = -4.14

Time can never be negative so t = 2.94 secs

ACCESS MORE
EDU ACCESS