Answer:
Acceleration, a =

t = 0.08 s
Solution:
As per the question:
The velocity of the player, v = 7.9 m/s
distance, d = 0.32 m
Now, consider the direction of the initial velocity of the player as positive and the acceleration to be constant:
Also, the final velocity of the player, v' = 0 m/s as he finally stops
Using the third eqn of motion:


a =

Also, From eqn (1) of motion:
v' = v - at
0 = 7.9 - 97.52t
t = 0.08 s