Answer
The maximum force on the ball from the player's foot is $~~4500~N$
Work Step by Step
$F(t) = (6.0\times 10^6)t-(2.0\times 10^9)t^2$
$F'(t) = (6.0\times 10^6)-(4.0\times 10^9)t$
To find the time $t$ when the maximum force occurs, we can let $F'(t) = 0$:
$F'(t) = 0$
$(6.0\times 10^6)-(4.0\times 10^9)t = 0$
$(6.0\times 10^6)=(4.0\times 10^9)t $
$t = \frac{6.0\times 10^6}{4.0\times 10^9}$
$t = 1.5\times 10^{-3}~s$
We can find the maximum force:
$F_{max} = (6.0\times 10^6)(1.5\times 10^{-3})-(2.0\times 10^9)(1.5\times 10^{-3})^2$
$F_{max} = (9000~N)-(4500~N)$
$F_{max} = 4500~N$
The maximum force on the ball from the player's foot is $~~4500~N$
