In a cricket match, the 0.16kg ball is bowled with a straight arm. during one particular delivery, the ball leaves the bowler's hand horizontally, with a speed relative to the ground of 80.8 mph. the bowler's arm is 0.64 m long and the bowler is running at 10.0 mph when the ball is released. calculate the centripetal force exerted on the ball in the bowler's hand, in n. note: 1 mph = 0.447 ms^-1.

412.1 N

Explanation:

First of all let's calculate the total speed of the ball as it is released, which is equal to the speed of the ball + the speed of the bowler:

Now let's convert it into m/s. We know that

So

And now we can calculate the centripetal force, which is given by:

where

m = 0.16 kg is the mass of the ball

v = 40.6 m/s is the speed of the ball

r = 0.64 m is the radius of the circular path (equal to the length of the bowler's arm)

Substituting: