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Let's start finding how high he jumped:
He started having a velocity of u0=2m/s. The acceleration of the gravity (g=9,8m/s2) slowed him down.
The time that his velocity became 0 (reached the highest point- point B) is:
u0-gtA->B=0 <=> 2-9,8*tA->B=0 <=> 9,8*tA->B=2 <=> tA->B=2/9,8 or about 0,2sec.
The distance he traveled upwards is: x1=u*tA->B-(1/2)atA->B2= 2*0,2 - (1/2)*9,8*(2/9,8)2 = 0,4 - 2/9,8 = 0,4 - 0,2 = 0,2m (it's rounded)
The swimmer now is 10,2 m above the water, and starts heading down.
He is accelerating at 9,8m/s starting with velocity=0. So:
Lets name tB->C the time needed to reach from point B to point C.
(x1+x2)=(1/2)g(tB->C)2 <=> 10,2 = (1/2)*9,8*(tB->C)2 <=>
tB->C2=2*10,2/9,8 <=>tB->C2=2.08 <=> tB->C=1.44sec (rounded)
Now, to find the velocity at point C:
uC=g*tB->C=9,8*1,44=14,11m/s (rounded)
Now,the swimmer is starting to accelerate upwards (slowing down because of the water) until he stops at point D.
the acceleration is 20m/s2. So, the time needed for him to stop is: uD=uC-a*tC->D <=> 0=14,11-20*tC->D <=>14,11=20*tC->D <=> tC->D=0,71sec (rounded)
And, the distance traveled: x3= uC*tC->D-(1/2)*a*tC->D2 = 14,11*0,71 - (1/2)*20*(0,71)2 = 10,02 - 10*0,5 = 5,02m
So far:
x1=0,2m x2=10m, x3=5,02m
uB=0 (at point B) uC=14,11m/s uD=0
At point B t=tA->B=0,2sec
At point C t= tA->B+tB->C=0,2+1,44=1,64sec
At point D t= tA->B+tB->C+tC->D=0,2+1,44+0,71=2,35sec
That proved quite lengthy... I hope that makes sense till now.. I'll now create and upload the graphs
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