Question

An object is thrown into the air going 9 m/s. How fast is it going 2 seconds later?-10.6 m/sO 10.6 m/s0 m/sO 0.8 m/s

Answer 1

To answer this question we have to remember that

$a=\frac{v_f-v_0}{t}$

In this case a=-9.81, t=2 and v0=9. Then

$\begin{gathered} -9.81=\frac{v_f-9}{2} \\ 2(-9.81)=v_f-9 \\ v_f=2(-9.81)+9 \\ v_f=-10.62 \end{gathered}$

Therefore the speed after two seconds is -10.62 m/s.