The bird perched on the swing has a mass of 52.0 g, and the base of the swing has a mass of 153 g. assume that the swing and bird are originally at rest and that the bird then takes off horizontally at 2.00 m/s. if the base can swing freely (i. e., without friction) around the pivot, how high will the base of the swing rise above its original level?

0.024m

Explanation:

The first thing you must do is to find the speed of the swing when the bird leaves. Use conservation of momentum (p) to do this. p(i) = p(f)

P(i) = 0 (no motion) = p(f) = m*v + M*V where m&v are the bird and M&V are the wing

So V = -m*v/M = -0.052 x 2/0.153= -0.68m/s

Now apply conservation of energy. SO K bottom = 1/2*M*V^2 = U (potential) = M*g*h

h= V^2/(2*g) = 0..68^2/(2*9.8) = 0.024m

increase masses, decrease separation