Leaving speed and distance after impact

[MechaMZ]

Homework Statement

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on the ground as a function of time. Starting from rest, a 61.0 kg athlete jumps down onto the platform from a height of 0.610 m. While she is in contact with the platform during the time interval 0 < t < 0.8 s, the force she exerts on it is described by the function below.
F = (9 200 N/s)t - (11 500 N/s2)t²

The Attempt at a Solution

(a) What impulse did the athlete receive from the platform?(981.333)
(b) With what speed did she reach the platform?(3.459)

(c) With what speed did she leave it?
(d) To what height did she jump upon leaving the platform?

my attempt for (c)
impulse = mv_f - mv_i
981.333 = 61(v_f) - 61(-3.459)
v_f = 12.628m/s

my attempt for (d)
V²_f = V²_i + 2ax
0 = 12.628² + 2(-9.81)x
x = 8.1277m

I don't understand why my answer for part c and part d are wrong =\

 

[willem2]

Gravity is still acting on the athlete during the 0.8s interval

 

[kuruman]

That's because you neglected to consider the weight of the athlete mg. The given force is also known as a normal force. Draw a free body diagram of the athlete and find an expression for the acceleration a(t) in terms of the normal force and the weight. Integrate to get the velocity. Your method for (d) is correct.