Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
Answer:
The speed of the ball is 9.07 m/s.
Explanation:
Given that,
Mass of the lead ball, m = 55 kg
Height of the tower, h = 55 m
We need to find the speed of the ball it has traveled 4.20 m downward, x = 4.2 meters
The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :
Here, a = g
v = 9.07 m/s
So, the speed of the ball is 9.07 m/s. Therefore, this is the required solution of given condition.