As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be


Answer:
vb = 22.13 m/s
So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.
Explanation:
In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:
K.E at A + P.E at A = K.E at B + P.E at B
(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)
(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)
where,
vₙ = velocity of roller coaster at point a = 0 m/s
hₙ = height of roller coaster at point a = 25 m
g = 9.8 m/s²
vb = velocity of roller coaster at point B = ?
hb = Height of Point B = 0 m (since, point is the reference point)
Therefore,
(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)
245 m²/s² * 2 = vb²
vb = √(490 m²/s²)
<u>vb = 22.13 m/s</u>
<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>
It makes no sense how you typed this problem out.
The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
The given parameters;
- initial horizontal velocity, vₓ = 16 m/s
- initial vertical velocity,

- time interval 1 seconds
The components of the velocity can be horizontal or vertical velocity.
The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.
The vertical component of the velocity is calculated as;

The horizontal component of the velocity is constant since it is not affected by gravity.
The horizontal component of the velocity = 16 m/s
Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
Learn more here:brainly.com/question/20349275