A ball of mass 10kg hits a smooth horizontal surface with a speed of `10sqrt(2)` m/s at angular of `45^(@)` as shown in the figure. The coefficient of resitution between the ball and the surface in 0.5 and the ball remains in contact with the surface for 0.1 sec. `vecF` is the instantaneous force extracted by the ball at any time t duringthe collison.

Options

A. The speed with the ball rebounds is `sqrt(125)` m/s
B. During collision `|vecF|` may be less than or equal to 1800N
C. During collision `|vecF|` may be greater than or equal to 100N
D. The average force exerted by ground collision is 1600N

Explanation

Applying impulse momentum equation for the3 ball
`rArrint_(0)^(0.1)(N-mg)dt=mv_(yf)-mv_(yi)`
`rArrint_(0)^(0.1)Ndt=10xx(10)/(2)+10xx10+10xx10xx0.1`
`rArrint_(0)^(0.1)Ndt=160rArr ltNgt=(intNdt)/(0.1)=1600`