Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Answer:
Position A/Position E
, 
Position B/Position D
,
, for 
Position C
, 
Explanation:
Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (
) is the sum of gravitational potential energy (
) and translational kinetic energy (
), all measured in joules. In addition, gravitational potential energy is directly proportional to height (
) and translational kinetic energy is directly proportional to the square of velocity.
Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:
Position A/Position E
, 
Position B/Position D
,
, for 
Position C
, 
A. because I had this question yesterday.
Answer:
c. Case iii
Explanation:
the ball will experience the largest change in case iii