Answer:
Acceleration, a = [tex]97.52 m/s^{2}[/tex]
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:
[tex]v'^{2} = v^{2} - 2ad[/tex]
[tex]0 = 7.9^{2} - 2a\times 0.32[/tex]
a = [tex]97.52 m/s^{2}[/tex]
Also, From eqn (1) of motion:
v' = v - at
0 = 7.9 - 97.52t
t = 0.08 s
