PHYSICS HELP please :(?

Refer to the figure below.
Take g = 9.8 m/s²
Take all downward quantities to be positive.

(a)
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u(y) = 0 m/s
Acceleration, a(y) = 9.8 m/s²
Time taken, t = 7.0 s

s(y) = u(y) t + (1/2) a t²
s(y) = (0) (7) + (1/2) (9.8) (7)²
Height of the cliff, s(y) = 240 m

(b)
Consider the vertical motion:
v(y) = u(y) + a(y) t
v(y) = (0) + (9.8) (7)
v(y) = 68.6 m/s

Final velocity of the rock
= √[v(x)² + v(y)²]
= √[20² + 68.6²]
= 71.5 m/s

Make an angle of tan⁻¹(68.6/20) = 73.7° below the horizontal line.

(c)
Consider the horizontal motion (uniform velocity motion):
Distance the rock travels horizontal before it hits the ground (range)
= v(x) t
= (20) (7)
= 140 m

a) The height of the cliff.
g = gravitational acceleration = 9.81 m/s²
viy = initial velocity in the vertical direction = 0 m/s
t = elapsed time of fall = 7.0 s
Δy = height of cliff (vertical distance) = to be determined

Δy = viy∙t + 0.5gt²
Δy = 0 m/s∙7.0 s + 0.5(9.81 m/s²)(7.0 s)²
Δy = 240.345 m = 240 m

b) The final velocity of the rock just before it hits the ground.
vfx = velocity in the horizontal direction = 20 m/s
vfy = final velocity in the vertical direction = to be determined
Vr = resultant final velocity = to be determined

vfy = viy + gt
vfy = 0 m/s + (9.81 m/s²)(7.0 s)
vfy = 68.67 m/s = 69 m/s

Vr = √[(vfx)² + (vfy)²]
Vr = √[(20 m/s)² + (69 m/s)²]
Vr = 71.52320533 m/s = 72 m/s

c) The distance the rock travels horizontally before it hits the ground.
vfx = velocity in the horizontal direction = 20 m/s
t = elapsed time of fall = 7.0 s
Δx = horizontal distance = to be determined

vfx = Δx / t
vfx∙t = Δx
Δx = vfx∙t
Δx = 20 m/s∙7.0 s
Δx = 140 m

y(t) = ½at² + v₀t  + y₀ is the basic formula you need to know.
Standard gravity is -9.80665 m/s²
For speed KE=PE, ie 1/2mv² = mgh
Here are the resolved Kinetics formulas: