A basketball is thrown upward from a height of 1.8 m at a velocity of 12 m/s. How long until it reaches the ground? Recall: h(t)=-4.9t^2-v_0 t+h_0, where v_0 is initial velocity and h_0 is initial height.
can anyone help?!

0.14secs

Step-by-step explanation:

Given the equation that models the height expressed as;

h(t)=-4.9t^2-v_0 t+h_0

Given v0 = 12m/s

h0 = 1.8m

The equation becomes

h(t)=-4.9t^2-12t+1.8

The height of the ball on the ground level is zero

Substituting into the equation

0 = -4.9t^2-12t+1.8

Multiply through by -10

49t²+120t-18 = 0

Factorize and get t:

t = -120±√120²-4(49)(-18)/2(49)

t = -120±√14400+ 3528/98

t = -120±√17928/98

t = -120±133.89/98

t = 13.89/98

t = 0.14secs

Hence it takes 0.14secs until it reaches the ground

i think c. but don't trust it.

step-by-step explanation:

15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

you.