Answer and Explanation:
The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence
For example let the initial height is 243 fit
After one bounce it will reach 243/3 =81 feet
After second bounce 81/3=27 feet
After third bounce 27/3 =9 feet
After fourth bounce 9/3 =3 feet
So a sequence is formed that is 243,81,27,9,3..........
Here 
Sum of infinite GP = 
From this formula we can find the total distance traveled by the ball
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:

with this value you can compute the frequency:
a)

b)
the mass of the block is given by the formula:

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

Finally, the amplitude is:

It is Tension as the other 3 answer choices would not make sense. Compression would mean you are pressing the rock on both sides or in this case, pushing it into the dirt. It can't be nuclear force as you are pulling out a rock. Air resistance would not make sense either as there is no air involved in the scenario at all.