A small pumpkin of mass 2.22 kg is dropped from rest from a height of 2.87 m. The pumpkin rebounds from the ground to a height of 1.12 m.

Question

a. With what speed does the pumpkin hit the ground?
b. With what speed does the pumpkin leave the ground?
c. What impulse was given to the pumpkin by the ground?
I know you use the rules of linear momentum but i just don't understand how to do this, any help please?

Answer 1

(1/2) m v^2 = m g h
so
v1 = -sqrt(2 g h) (coming down)
h = -2.87 coming down

going up same deal but opposite sign for v
v2 = + sqrt(2*9.8*1.12)

change in momentum = m (v2-v1)