The angle the rope make with the vertical when the ball's kinetic energy is halved is 14.78°.
Let K = 1/2mv² be the balls initial kinetic energy when it is pushes horizontally. It rises to a height, h has potential energy U = mgh and has kinetic energy half of its initial kinetic energy. So, K' = 1/2K.
So, from the law of conservation of energy,
K = U + K'
K = U + 1/2K
K - 1/2K = U
K/2 = U
1/2(1/2mv²) = mgh
1/4mv² = mgh
1/4v² = gh
h = v²/4g
Since v = initial speed of ball = 1.3 m/s and g = acceleration due to gravity = 9.8 m/s².
Substituting the values of the variables into the equation, we have
h = v²/4g
h = (1.3 m/s)²/(4 × 9.8 m/s²)
h = 1.69 m²/s² ÷ 39.2 m/s²
h = 0.043 m.
Since the length of the rope l = 1.3 m and the distance of the ball from the pivot when it is h metres above its lowest point is L = l - h = 1.3 m - 0.043 m = 1.257 m.
So, the angle the rope makes with the vertical is Ф and cosФ = L/l
= 1.257/1.3
= 0.9669
Ф = cos⁻¹(0.9669)
Ф = 14.78°
So, the angle the rope make with the vertical when the ball's kinetic energy is halved is 14.78°.
Learn more about oscillatory motion here:
https://brainly.com/question/20519190
