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=(v_f-v_0)/(t)`

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

`\begin{gathered} -9.81=(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.