A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.

a. find the momentum of the ball of dough at its maximum height.

b. find the momentum of the ball of dough halfway to its maximum height on the way up

Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh = > h = v^2/2g h = 15^2/2x9.8 = 11.48m = > Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) = > V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V = > M = 0.10 x 10.81 = > M = 1.081kg-m/s