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

the answer is 37

step-by-step explanation: