Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.
Answer:
Probability of tunneling is 
Solution:
As per the question:
Velocity of the tennis ball, v = 120 mph = 54 m/s
Mass of the tennis ball, m = 100 g = 0.1 kg
Thickness of the tennis ball, t = 2.0 mm = 
Max velocity of the tennis ball,
= 89 m/s
Now,
The maximum kinetic energy of the tennis ball is given by:

Kinetic energy of the tennis ball, KE' = 
Now, the distance the ball can penetrate to is given by:


Thus



Now,
We can calculate the tunneling probability as:



Taking log on both the sides:


If you were given distance & period of time, you would be able to calculate the speed.
Hope this helps!
It is Real,Virtual,The Same Size, Inverted