physics help please!?

(a)
Take g = 9.81 m/s

Speed that the diver leaving the cliff, u
= Speed that the diver runs towards the edge
= (45.0 m) / (6.40 s)
= 7.03 m/s

Consider the vertical motion (uniform acceleration motion, downward quantities to be positive) :
Initial speed, u = 7.03 sin19.0° m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 24.0 m

s = ut + (1/2)at²
24.0 = (7.03 sin19.0°)t + (1/2)(9.81)t²
(1/2)(9.81)t² + (7.03 sin19.0°)t - 24.0 = 0
t = {-(7.03 sin19.0°) ± √[(7.03 sin19.0°)² + 4(1/2)(9.81)(24/0)]} / [2(1/2)(9.81)]
t = 1.99 s or t = -2.46 s (rejected)
Time taken = 1.99 s


(b)
Horizontal distance
= (7.03 cos19.0° m/s) × (1.99 s)
= 13.2 m

The plan:
i) Find his velocity down the slope as he runs towards the edge.
ii) Resolve that initial velocity into its vertical and horizontal components
iii) Kinematics on the vertical motion to find the time taken to hit the water
iv) Use the time (t) fron (iii) with the constant horizontal component from (ii) to find horizontal distance.

i) Velocity down slope = displacement/time = (45.0 m / 6.40 s) = 7.03 m/s

ii) Vertical component of launch speed = 7.03sin(19.0°) = 2.29 m/s
Horizontal component of launch speed = 7.03cos(19.0°) = 6.65 m/s

iii) Kinematics on the vertical component of motion. The symbols used vary from place to place. I use the “suvat” symbols: s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time

I choose arbitrarily to say that "downwards is the positive direction". What do we know here?
s = 24.0 m
u = 2.29 m/s
v = {not interested}
a = 9.81 m/s² {Acceleration due to gravity downwards}
t = ? {That is what we want to know}

Equation choice: I could use s = ut + ½at² to find (t) directly, but I am too full of Sunday lunch to fool about with quadratics, so I'll find (v) first...
v² = u² + 2as
v² = 2.29² + 2 * 9.81 * 24.0
v² = 476.12
v = 21.8 m/s

Now: v = u + at
21.8 = 2.29 + 9.81t
(a) t = 1.99 seconds

iv) So travelling with a horizontal component of 6.65 m/s, how far does he get in the 1.99 s it takes to hit the water?

(b) distance = speed * time = (6.65 m/s * 1.99 s) = 13.2 m

"cliff diver stands 45.0 m from the edge of a cliff...reaches the edge of the cliff in a time of 6.40 s."
That tells you his initial velocity v = distance/time.

And you're told it's at an angle θ relative to the horizontal. So you know his horizontal velocity vx = v cos(θ) and his initial vertical velocity vy = v sin(θ).

a) Accelerated motion, d = vy * t + (1/2)at^2. You know d, vy, and a. Solve for t. Note that d and a are downward and vy is upward, so you should use positive signs for one and negative for the other. Doesn't matter whether you use positive for up or for down, just be consistent.

b) Constant-velocity motion, d = vx * t.