Mass = m, Radius: R . Sphere is solid.
h = 4R/5 above the horizontal center line of the sphere. blow horizontally.
I = impulse = F * Δ t = Δp as per the conservation of linear momentum.
Linear Momentum of sphere after impulse = m v = p = I
As h > 2 R / 5 , The ball probably slips along with rolling. So, perhaps v ≠ R ω. If the horizontal impulse is given at a height 2/5 R, for a solid sphere, then it rolls without slipping.
Moment of inertia MI about cm = 2/5 * m R²
Angular momentum = MI * ω = 2/5 m R² ω
L = h * p = Angular impulse given, as initial angular momentum = 0
=> h * p = 4/5 R * m v = 4/5 m v R = 2/5 m R² ω
=> ω = 2 v / R and v = R ω / 2 ---- (2)
So ω = 2 I / (m R)
We need to find the Time duration for the ball to rotate by π radians, 180 deg.:
= t = π / ω = π m R / (2 I)
So the horizontal (translational, linear) displacement
= s = v * t = I / m * π m R / (2 I) = π/2 * R
Also, the Energy K.E. given by the impulse:
½ m v² + ½ I ω² = ½ m v² + ½ * 2/5 * m R² * ω²
= m [v² / 2 + R² ω² /5]
= m R² ω² [1/8 + 1/5 ]
K. E. = 13/40 * m R² ω²
= 13/10 * m v²