Physics helps plsease

A projectile, red with unknown initial velocity, lands 22:0 s later on the side of a hill,
3270 m away horizontally and 445 m vertically above its starting point. (Ignore any
eects due to air resistance.)
(a) What is the vertical component of its initial velocity?
(b) What is the horizontal component of its initial velocity?
(c) What was its maximum height above its launch point?
(d) As it hits the hill, what speed did it have and what angle did its velocity make
with the vertical?

A projectile, red with unknown initial velocity, lands 22:0 s later on the side of a hill, 3270 m away horizontally and 445 m vertically above its starting point. (Ignore any effects due to air resistance.)

In the 22.0 seconds, the projectile moves horizontally 3270 m
Horizontal displacement = v * cos θ * time

3270 = v * cos θ * 22

Eq. #1
v * cos θ = 3270 ÷ 22 = horizontal component of its initial velocity

Vertical displacement = v * sin θ * time – ½ * g * time^2
445 = v * sin θ * 22 – ½ * 9.8 * 22^2
445 = v * sin θ * 22 – 2371.6
add 2371.6 to both sides

2816.6 = v * sin θ * 22

Eq. #2
v * sin θ = 2816.6 ÷ 22 = vertical component of its initial velocity

Divide Eq #2 by Eq #1
tan θ = 2812.6 ÷ 3270

θ = 40.74°

v * cos 40.74° = 3270 ÷ 22
v = 196.2 m/s

v * sin 40.74 = 2816.6 ÷ 22
v = 196.2 m/s OK

At the maximum height, vertical velocity = 0
Distance = average velocity * time
Average velocity = ½ (v * sin θ + 0)
Time to reach maximum height = v * sin θ ÷ 9.8

Maximum height = (½ * v * sin θ) * (v * sin θ ÷ 9.8) = (v * sin θ)^2 ÷ 19.6

Maximum height = (196.2 * sin 40.74)^2 ÷ 19.6

Final vertical velocity = Initial vertical velocity – 9.8 * total time
Final vertical velocity = 2816.6 ÷ 22 – 9.8 * 22 = -87.57 m/s

Final horizontal velocity = Initial horizontal velocity = 3270 ÷ 22 = 148.64 m/s