The speed at which the ball bounced back up after hitting the floor (v1) is 3.96 m/s. The ball lost 20% of its energy after the first bounce, predicting the next maximum height (h2) at 0.64 m. The speed after the second bounce (v2) will be less than v1, and the time to reach the peak height (T2) will also be less than the time after the first bounce (T1).
To find the speed (v1) at which the ball bounced back up after hitting the floor, we can use the formula for the velocity of an object just before impact or just after rebounding: v = √(2gh), where g is the acceleration due to gravity (9.8 m/s2), and h is the height.
Using h1 = 0.8 m gives v1 = √(2 * 9.8 * 0.8) = √(15.68) ≈ 3.96 m/s.
The percent of energy loss to heat, air resistance, and the deformation of the ball can be found by comparing the initial and final heights.
The energy loss is (1 - (h1/h)) × 100%, which gives us (1 - (0.8/1.0)) × 100% = 20%.
To predict the next maximum height (h2), if the loss of energy is constant, we consider the 20% loss.
Therefore, h2 = h1 × (1 - 0.2) = 0.8 m × 0.8 = 0.64 m.
As for how v1 compared to v2, since the height decreases, the speed after the second bounce will be less than v1.
The time of ascent to the peak height for both bounces (T1 and T2) can also be compared.
With a constant loss of height and respective speed, T2 will be less than T1 because T = √(2h/g).