Please answer... physics problem....Do NOT round plz?

Question

A 51.0 kg diver steps off a diving board and drops straight down into the water. The net force when in the water is an upward 1020 N. If the diver comes to rest 5.0 m below the water's surface, what is the total distance between the diving board and the diver's stopping point underwater?

Answer 1

using v^2=u^2+2as (where s is the drop)

v^2=0+2gs
v^2=2gs (where v is the speed he hits the water)

when in the water the upwards force = 1020N
so(using F=ma)
1020=51a
a=1020/51 (ie his acceleration upwards)

back to v^2=u^2+2as

0=2gs-2(1020/51)*5 (negative because acceleration is up)
s=10.2m

Total distance is 15.2m

Answer 2

First, determine the deceleration of diver as he impacts the water.
F = ma
F was given 1020N, m was given 51kg.
so a =F/m = 1020/51 = 20 m/s^2.

Next, determine the time from impact of water to the diver's rest point below water surface.

X = 0.5 a t^2
X was given 5.0m
5 = 0.5 * 20 * t^2
t^2 = 0.7071...

Now, determine the speed of diver at impact
v = at = 20 * 0.7071... = 14.142...

Next determine the time he took from leaving the diving board to impact in water.
Let us use g, Earth's gravitional constant, as 9.8 m/s^2.
v = gt
14.142 = 9.8 * t
t = 1.443... (seconds)

Using the same formula again, determine the height of the diving board
X = 0.5 a t^2, where a = g
X = 0.5 * 9.8 * (1.443... )^2
X = 10.204 m

So total distance = 5.0 + X = 15.204m

Answer 3

Try an approach using energy. You don't really expect us to do this tedious question for you, do you?