Physics(Please help)
posted by Hannah on .
1) The velocity of a diver just before hitting the water is 11.8 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 1.06 s of the dive?
I know that the formula for displacement is deltaX= x  x0. Would I do 11.8  1.06?

Use the acceleration of gravity to find her velocity 1.2 s before impact. Average the two velocities times the 1.2 s to get the displacement.
v = v0  g*t
10.4 = v0  9.81*1.2
v0 = 1.372
So the diver was above the level from which she dove from at 1.2 s before impact. The time of max height can be found by using v0 = 0 in the above equation.
t = v/g = 10.4/9.81 = 1.06 s
So max height was reached 1.06 seconds before impact.
Vavg = (1.372  10.4)/2
Vavg = 4.514
Displacement = Vavg*1.2 = 5.4 m
So her displacement during the last 1.2 s of the dive is 5.4 meters. Displacement is just the distance from starting point to finishing point regardless of the path taken. In this case her initial motion was upward to max height and then downward to the water. But the positive distance upward is irrelevant to the displacement in this case. 
For some reason when i put in 5.4 as the answer it said I was incorrect. I do not know why.

The diver’s motion is free fall.
Therefore
h =g•t²/2
v=g•t
The speed 11.8 m/s was gained for the time
t = v/g = 11.8/9.8 = 1.2 s.
The height from which the diver janped was
h = g•t²/2 =9.8•1.2²/2 = 7.1 m.
The time for the first part covering is
t1= 1.2  1.06 =0.14 s.
The distance for this time is
h1 = g•t1²/2 = 9.8•0.14²/2 = 0.1 m.
Δh =7.1 0.1 = 7 m