The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
With constant angular acceleration
, the disk achieves an angular velocity
at time
according to

and angular displacement
according to

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

b. Under constant acceleration, the average angular velocity is equivalent to

where
and
are the final and initial angular velocities, respectively. Then

c. After 1.00 s, the disk has instantaneous angular velocity

d. During the next 1.00 s, the disk will start moving with the angular velocity
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to

which would be equal to

Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒
Answer: The box was moving with a velocity of 0.256m/s when it hit the spring
Explanation: Please see the attachments below