A 0.400 kg ball is shot directly upward at initial speed 40.0 m/s. What is its angular momentum about a point 2.0 m horizontal 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 the point 2m from the horizontal launch point due to the gravitational force when the ball is c) at maximum height and d) halfway back to the ground?

I know that because max height is v=0, there is 0 angular momentum, but i can't figure out the rest

5 answers

Do the second question first. (It is worded incorrectly I think because there is in fact no torque on the ball since the force goes right through its center. There is however a torque about that point 2 meters from the launch point) The torque is the same at the top and halfway down because the perpendicular distance from the force vector to the launch point is always 2 meters. The magnitude of the torque = m g (2)

Well, we better figure out how fast it is going when it is halfway down. You already answered for at the top, zero because velocity is zero.
a = - 9.8
v = 40 - 9.8 t
h = 0 + 40 t - 4.9 t^2
when h = 20
20 = 40 t - 4.9 t^2
4.9 t^2 - 40 t + 20 = 0
t = (1/9.8)[ 40 +/- sqrt(1600 -392)]
t = (1/9.8) [ 40 +/- 34.8 ]
t = 7.63 on the way down (the plus sign)
v = 40 - 9.8(7.63) = -34.8 m/s
angular momentum = m V x R
= m (-34.8)(horizontal distance which is 2)
= -69.6 * mass

THANK YOU SO MUCH!

Torque is always (2m)xF=-2mg
velocity at h/2 is -vo/root2=-28.284
So Angular momentum is
(2m)xmv=-22.627

asd