A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If air resistance does –1.8 J of work on the ball, how much energy goes into heating the ground and the ball when it bounces?

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

Let g = 9.81 m/s2

Since 1.8J of 2.51 J is due to work by air resistance, the rest of the energy (2.51 - 1.8 = 0.71 J) is would go to heating in the ground and ball when it bounces.

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

snell's law

refractve index = sin incidence / sin refraction