Question

In the free fall experiment, we dropped a ball with a mass of 0.05kg from a height of 20cm. Using the stopwatch, the time period for its fall became 0.22 sec, so the final movement energy is equal to:

Answer 1

The initial velocity V = 0

And the acceleration is due to gravity is g = 9.8 m/s²

So the time taken by ball to reach ground

From newton's second equation

Y =.2= gt²/2

So the t= √.4/9.8 = .2 sec

And the velocity at ground v = √2gh (from third equation)

V = √(2*9.8*.2) =1.98 m/s

Now the time remais t = .22-.2 = 0.2 sec

So for. 2 sec it will jump up due to reaction force by earth

So hight traveled by ball after collision ground

Y = 1.98*.02 - 9.8*(.02) ²/2 = .0415 m =4.15 cm

So from the energy conservation campairs the energy

At hight 20 cm (rest) and hight 4.15 cm

E_1 = only potential energy

E_2 = movement energy and potential energy

So E_1= E_2

mg*.2 = mg*.0415+movement energy

So the moment energy

= mg*(.2-.0415) = .05*9.8*1.585= . 776 joules

E = . 776 joules