Question

I have a homework problem that I could use some help with if anyone knows the answer.
"A ball is thrown upward with a speed of 22m/s. What is the ball's height after 2s"

Answer 1

S = VoT + (1/2)gT^2

where

S = ball's attained height
Vo = 22 m/sec.
T = 2 sec.
g= 9.8 m/sec^2

Substituting

S = 22(2) + (1/2)(-9.8)(2^2)

Answer 2

the negative sign (-9.8 m/sec^2). This simply implies that the ball is being slowed down by gravity as it is going up.

Answer 3

Cool. Thanks a lot!