A240 g ball is dropped from a height of 2.1 m, bounces on a hard floor, and rebounds to a height of 1.6 m. the figure shows the impulse received from the floor. what maximum force does the floor exert on the ball?
impulse= .005*fmax

FΔt=mΔv−−>F=m(Δv)/(Δt)
We can find the speed by calculating the potential energy and converting it to kinetic energy. Ep=Ek−−>mgh=mv^2/2

v=(√2gh)=(√2∗9.81∗2.1)≈6.42m/s

so then F=.240∗6.42/0.005=308 N (we know the time because of the above mentioned impulse equation.)