All credit goes to the person who answered this years ago:)
Reaches max height at t = 2.42s.
Step-by-step explanation:
I've assumed we are neglecting air resistance. If not let me know and I'll update.
We want to examine the behaviour of the ball in the y-direction. In the absence of air resistance the only force acting on the ball is gravity, which produces an acceleration in the negative y direction. From Newton's 2nd law:
m
d
2
y
d
t
2
=
−
m
g
m
d
2
y
d
t
2
=
−
m
g
Integrating:
d
y
d
t
=
−
∫
g
d
t
d
y
d
t
=
−
g
⋅
t
+
C
From initial conditions,
d
y
d
t
=
v
y
(
t
)
,
v
0
=
v
(
0
)
=
31
⋅
sin
(
50
)
∴
v
y
(
t
)
=
v
0
−
g
⋅
t
The maximum height will be reached at
v
y
=
0
so we solve for t.
v
y
(
t
h
max
)
=
0
⇒
v
0
=
g
⋅
t
h
max
t
h
max
=
v
0
g
=
31
⋅
sin
(
50
)