Acceleration is -10 NOT 9.8

Question

4. An orange is tossed upward at 21m/s. What is the velocity of the orange 3.5s later? What is the height of the orange at this time? Is the orange still travelling up or is it travelling down?

Answer 1

The equation of motion is s(t) = s₀(t) + v₀(t)t - gt².

The velocity v(t) = s'(t) = v₀(t) - 2gt.

Your question seems to indicate that you want to use 10 for g, the acceleration due to gravity.

v₀(t) = 21 and you want to know v(3.5), s(3.5)and the sign of velocity at t=3.5.

I will not due the substitutions for you, especially since it is unclear why you would want to change the value of g which is a constant of nature.