Could I get some help with the problem, please?

Question

A baseball is thrown upward with an initial velocity of 88 feet per second (𝑣0 = 88). When will the ball be 120 feet above the ground? [ℎ(𝑡) = −16𝑡 2 + 𝑣0𝑡].

Answer 1

v₀=88ft/s

h=120ft

g=32 ft/s²

h=v0*t-0.5*g*t²

120ft=88ft/s*t-16ft/s²*t²

16t²-88t+120=0

a=16

b=-88

c=120

Divide the entire equation by four

a=4

b=-22

c=30

(-b+sqrt(b²-4ac))/2a or (-b-sqrt(b²-4ac))/2a

(22+sqrt(484-480)/8 or (22-sqrt(484-480))/8

(22+2)/8 or (22-2)/8

t=3 seconds on way down and at 2.5 seconds on the way up.

t=2.5 seconds and 3.0 seconds

Answer 2

-16t²+88t = 120

-16t²+88t -120 =0

Need to solve for t.

Divide by -4

4t²-22t+30=0

t = (22±√(484-480))/8

t = 24/8 , 20/8

t = 3, 2.5

Since the path is an inverted parabola, the first time it it reaches 120 feet going up is 2.5 seconds and coming down is at t = 3 seconds.