a pitcher throws a baseball straight up with an initial speed of 27m/s.?

Question

a) How long does it take the ball to reach its highest point?
b)how high does the ball rise above it's release point?

Answer 1

17.04 seconds. 460 meters.

Answer 2

using the frst eqn of motion
v = u + at
v = 0 as vel at highest pt. = 0
therefore
0 = 27 + -10(Acc is negative)
=
27/10 = t
=
sumthing like 2.7...
Tie = 2.7 approx..
Therefore
s = ut+1/2at^2
or v^2 = u^2 +2as
=
-27*27/2*10 = s
=
72.9/2
=
36.45 Metres