Question

In the absence of air resistance, a ballplayer tosses a ball
straight up.
(a) By how much does the speed of the ball decrease
each second while it is ascending?
(b) By how much does its speed increase each second
while it is descending?
(c) How does the time of ascent compare with the time
of descent?

Answer 1

a. V = Vo-g*t
V = Vo-9.8m/s^2*1s
V = Vo-9.8m/s. s.

The speed decreases by 9.8 m each second.

b. V = Vo + g*t
V = Vo + 9.8m/s^2*1s
V = Vo + 9.8m/s.
The speed increases by 9.8 m each second.

c. V = Vo + g*t = 0 @ max. ht.
Vo + g*t = 0
Vo - 9.8t = 0
9.8t = Vo
Tr = Vo/9.8 = Rise time=Time of ascent.

V = Vo + g*t
V = 0 + 9.8*t
Tf = V/9.8
V = Vo during ascent.
Therefore, Tf = Vo/9.8 = Tr.