A 0.13 kg ball of dough is thrown straight up into the air with an initial speed of 18 m/s. The acceleration of gravity is 9.81 m/s2 .What is its momentum halfway to its max- imum height on the way up?

Answer 1

hmax = (V^2-Vo^2)/2g
hmax = (0-18^2)/-19.6 = 16.53 m.

V^2 = Vo^2 + 2g*h
V^2 = 18^2 - 19.6*(16.53/2) = 162
V = 12.73 m/s.

Momentum = m*V = 0.13 * 12.73 = 1.65
kg-m/s.

Answer 2

Henry did it wrong, you have to divide Hmax by two after finding Hmax. Then you would continue doing the same steps