Center of Mass and Collision

To determine when the ball will hit the surface for the second time, we start by calculating the velocity just before it hits the surface for the first time. Using the equation for the velocity of a freely falling object, the velocity is:

$ v = \sqrt{2 \times g \times h} = \sqrt{2 \times 9.8 \times 4.9} = 9.8 \, \text{m/s} $

where:

$ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity)

$ h = 4.9 \, \text{m} $ (height)

Next, calculate the velocity after the first bounce using the coefficient of restitution $ e = \frac{3}{4} $:

$ v_{\text{bounce}} = e \times 9.8 = \frac{3}{4} \times 9.8 = 7.35 \, \text{m/s} $

The time taken for the ball to travel up and back down after the first bounce, until it hits the surface again, is:

$ t = \frac{2v_{\text{bounce}}}{g} = \frac{2 \times 7.35}{9.8} $

$ t = \frac{14.7}{9.8} = 1.5 \, \text{s} $

Therefore, the ball will hit the surface for the second time after 1.5 seconds.

According to the law of conservation of linear momentum at the highest point,

$mv\cos \theta = {m \over 2}(v\cos \theta ) + {m \over 2}v'$

$ \Rightarrow mv\cos \theta - {m \over 2}v\cos \theta = {m \over 2}v'$

$ \Rightarrow v' = 2\left( {v\cos \theta - {{v\cos \theta } \over 2}} \right) = v\cos \theta $