Answer:
Part a)

Part b)

Explanation:
Part A)
As we know that time period of the motion is given as

so we have



now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

so we have



Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have



so now we have


Explanation:
Below is an attachment containing the solution.
The answer would be 2.63. Your welcome. This has been changed to the correct answer.
Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
Answer:
The option is B is not true for Hubble telescope.