Question

A player kicks a 0.5 kg foitball and gives it a velocity of 10 m/ s starting from rest the contact between the foot and the ball lasts for 1/50 sec. What is the force of impact?

Answer 1

$$\begin{gathered} massofthefootballm=0.5Kg. \\ changeinvelocity=v-u=10-0=10m/s. \\ timetaken=1/50second.thereforeforceofimpact, \\ F=\frac{m\left(v-u\right)}{t}=\frac{0.5\left(10-0\right)}{1/50}=5\times 50=250N. \end{gathered}$$