Question III

III) A 0.63 kg ball is thrown up with an initial speed of 14 m/s and reaches a maximum height of 8.1 m. How much energy is dissipated by the air drag acting on the ball during the ascent? (10 pts). How much higher would it have gone were the air drag negligible? (10 pts).

Solution:
The energy that the ball has initially is
$E_{i} = \frac{1}{2}mv_{i}^{2}$
E_i = 61.7 J
The potential energy that the ball has at its maximum height is
E_f = mgh
E_f = 50.0 J
So, the change in energy due to drag is given by
$W_{drag} = \Delta E ~=~ E_{f} - E_{i}$
W_drag = -11.7 J
If the full 61.7 J of energy was converted into potential energy, then the ball would have risen to a height given by
mgh = 61.7 J
h = 10.0 m