Question

A projectile is shot on level ground with a
horizontal velocity of 21 m/s and a vertical
velocity of 45 m/s.
Part 1) Find the time the projectile is in the air.
The acceleration due to gravity is 9.8 m/s^2 .Answer in units of s.
Part 2) Find the range R. Answer in units of m.
Part 3) Let the magnitude of v0 be the same as in part 1, but consider the angle θ required to obtain the maximum range.
Find the height attained for the maximum range.Answer in units of m.

For my answer this what but it seem to be the wrong answer. Please help me.

1)
d = Viy t + 1/2 g t^2
d = 0
0 = 45 t - 4.9 t^2
4.9 t = 45
t = 9.2 s

2)R = X = Vx t = 21(9.2) = 193 m

3)Vo^2 = 21^2 + 45^2 = 246.6
Vo = 49.66 m/s or about 50
then at 45°
Vx = Vo sin45 = 49.66 sin45 = 35.1 or 35 m/s
d = Vyi t + 1/2 g t^2
0 = 35 t - 4.9 t^2
4.9 t = 35
t = 7.1 s

Answer 1

Xo = 21 m/s.
Yo = 45 m/s.

1. Y = Yo + g*Tr = 0 at max.ht.
Tr = -Yo/g = -45/-9.8 = 4.59 s. =
Rise time.
Tf = Tr = 4.59 s. = Fall time.
Tr+Tf = 4.59 + 4.59 = 9.18 s. = Time in air.

2. R = Xo*(Tr+Tf) = 21m/s * 9.18s
= 193 m.

3. A = 45o for max range.
Vo = Sqrt(21^1+45^2) = 49.7 m/s[45o].

Yo = 49.7*sin45 = 35.11 m/s.

Y^2 = Yo^2 + 2g*h = 0.
h = -(Yo^2)/2g = -(35.11)^2/-19.6
= 63 m.