Show all work! I don't understand this! Brainleist!
I attached a picture this time!

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