a 0.400 kg ball is shot directly upward at initial speed 40.0 m/s. what is its angular momentum about p, 2.00 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? what is the torque on the ball about p due to the gravitational force when the ball is (c) at maximum height and

The maximum height of the ball is 80m.We are given that,Mass of the ball = m =0.400kgInitial speed of the ball = u = 40.0m/s²Distance cover by the ball = r = 2.00mTherefore, when the ball shot upward then the maximum height can be calculated by,v² - u² = 2ghWhere, v is final velocity , u is the initial velocity , g is the gravity i.e. 10m/s², h is height of the ball.Putting the values in above equation of motion, here final velocity of the ball is zero,0 - (40×40) = 2×10×hh = 80mTherefore, the maximum height of the ball would be 80m.To know more about maximum heighthttps://brainly.com/question/14062716#SPJ4