Answer:
7.08 m/s²
Explanation:
Given:
v₀ = 20.0 m/s
v = 105 m/s
t = 12.0 s
Find: a
v = at + v₀
105 m/s = a (12.0 s) + 20.0 m/s
a = 7.08 m/s²
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:


Answer:
meters
Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function
The maxima of the function occurs when the slope is zero. i.e.

Hence the maxima occurs at t=1.63 seconds
The maximum value of f is

hence maximum height is found to be
meters