Answer:
football player decceleration is 70.01 m/s^2
Explanation:
given data:
x = 0.360
initial velocity vi = 7.10m/s
final velocity vf = 0 m/s
we know that acceleration is given as a
[tex]a =\frac{ vf^2 -vi^2}{2x}[/tex]
putting all value to get the desired value of acceleration:
[tex]a = \frac{ -(7.10)^2}{2*.360}[/tex]
a = - 70.01 m/s^2
negative sign indicate deceleration
football player decceleration is 70.01 m/s^2
