An electromagnet on the ceiling of an airplane holds a steel ball. When a button is pushed, the magnet releases the ball. The ex
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The motion of the ball on the vertical axis is an accelerated motion, with acceleration
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
(1)
Now let's assume that
is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 
. So the maximum height of the second ball is
(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
Now let's assume that
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find