Question

4) Use the Energy-Interaction Model to show that an object thrown vertically upward will have the same speed going up and coming down at the same height. a) Draw a picture of the initial and final states. b) Apply two methods of solving this problem: i) Construct two energy system diagrams. One diagram should be from the point of release of the ball to some intermediate height as the ball is traveling upward, less than the maximum height. The second diagram should be from the point of release of the ball to that same intermediate height as the ball is on its way down. Then, compare the two diagrams. ii) Construct one energy system diagram using the whole interval as drawn in a). What do you conclude about the changes in energy systems? c) Can you think of a condition where the speed going up and coming down at the same height would not be the same? Will the speed be lower or higher on the way down?

Answer 1

C. in C part if there is air resistance acting F= -bv

where v is velocity

then it can happen
Energy while throwing the ball rightarrow 1/2 mv^2_0 + mgh -ground energy while ball reaches ground + 1/2 mv^2_1 + mgh Since energy is conserved Since h is same = 1/2 mv^2_0 = 1/2 mv^2_1 = > v_1 = v_0