If he is jumping you are adding force which means that you will be falling twice as fast
Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

Where,
T = Tension
Linear density
Our data are given by
Tension , T = 70 N
Linear density , 
Amplitude , A = 7 cm = 0.07 m
Period , t = 0.35 s
Replacing our values,



Speed can also be expressed as

Re-arrange to find \lambda

Where,
f = Frequency,
Which is also described in function of the Period as,



Therefore replacing to find 


Therefore the wavelength of the waves created in the string is 3.49m
Answer:
it will show a continuous rise in value. The rise will be sinusoidal.
Explanation:
a. 27
after that all you need to do is use the work formula as you will be able to acheive results