a.
The object's speed at 2.20 m below balcony level is 8.74 m/s
Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.
Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and
So, E = E'
U + K + f = U' + K' + f'
where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).
So,
U + K + f = U' + K' + f'
mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh
mgh = mgh' + 1/2mv² + 0.10mgh
Dividing through by m, we have
gh = gh' + 1/2v² + 0.10gh
So, gh - 0.10gh = gh' + 1/2v²
0.90gh = gh' + 1/2v²
1/2v² = 0.90gh - gh'
1/2v² = g(0.90h - h')
v² = 2g(0.90h - h')
Taking square-root of both sides, we have
v = √[2g(0.90h - h')]
where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the values of the variables into the equation, we have
v = √[2g(0.90h - h')]
v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]
v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]
v = √[2 × 9.8 m/s²(3.901 m)]
v = √[76.4596 m²/s²]
v = 8.74 m/s
So, the object's speed at 2.20 m below balcony level is 8.74 m/s
b.
Yes it does matter when we apply 10% loss before V calculations
We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.
So, yes it does matter when we apply 10% loss before V calculations
Learn more about speed of a falling object here:
https://brainly.com/question/18732868
