Question

A 0.500 kg object is thrown vertically upward with an average applied force of 8.20 N by a student. The force is applied through a displacement of 1.50m.

a. What is the average net force acting on the object?

b. What is the velocity of the object when it leaves the student's hand? (Assume initial velocity is zero)

Answer 1

8.2 UP and .5(9.81 ) down
= 3.295 Newtons up

Work done
=Force * 1.5 meters = 3.295*1.5
= 4.9425 Joules
so
(1/2) (.5) v^2 = 4.9425

v = 4.45 m/s