50. With

+

y upward, we have y

0

= 36.6 m and y = 12.2 m. Therefore, using Eq. 2-18(the last equation in

Table 2-1), we find

y

− y

0

= vt +

1

2

gt

2

=

⇒ v = −22 m/s

at t = 2.00 s. The term speed refers to the magnitude of the velocity vector, so the answer is

|v| =

22.0 m/s.