A point mass is shot vertically up from ground level with a velocity of 4 m/s at time, t = 0. It loses 20% of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is 10 m/s² and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of

The correct option is 4.

1. Understanding the Motion:
A point mass is projected vertically upwards from ground level with an initial velocity of $u=4\text{m/s}$. Under the action of gravity, it goes up to its maximum height and then falls back to the ground. Upon hitting the ground (the first collision), it loses 20% of its impact velocity, which means its coefficient of restitution is:
$$e=1-0.20=0.8$$

2. Time Taken for Each Bounce:
The time taken to reach the maximum height in the initial journey (before the first collision) is:
$$t=\frac{u}{g}=\frac{4}{10}=0.4\text{s}$$

Since the time of ascent equals the time of descent in the absence of air resistance, the time of flight for the first motion is:
$$T_1=2t=2\left(0.4\right)=0.8\text{s}$$

After the first collision, the rebound velocity is:
$$u^{\text{'}}=e·u=0.8\times 4=3.2\text{m/s}$$

The time of ascent for this second phase is:
$$t^{\text{'}}=\frac{u^{\text{'}}}{g}=\frac{3.2}{10}=0.32\text{s}$$

So, the time of flight for the second bounce is:
$$T_2=2t^{\text{'}}=2\left(0.32\right)=0.64\text{s}$$

After the second collision, the rebound velocity is:
$$u^{\text{'}\text{'}}=e·u^{\text{'}}=0.8\times 3.2=2.56\text{m/s}$$

The time of ascent for this third phase is:
$$t^{\text{'}\text{'}}=\frac{u^{\text{'}\text{'}}}{g}=\frac{2.56}{10}=0.256\text{s}$$

So, the time of flight for the third bounce is:
$$T_3=2t^{\text{'}\text{'}}=2\left(0.256\right)=0.512\text{s}$$

3. Calculating the Total Time:
As visible in the provided analysis image, the times of ascent $t,t^{\text{'}},t^{\text{'}\text{'}},...$ form a geometric progression (GP) with a first term of $a=0.4\text{s}$ and a common ratio of $r=0.8$.

The total time before coming to rest is the sum of the times of flight for all bounces:
$$T_{\text{total}}=2(t+t^{\text{'}}+t^{\text{'}\text{'}}+...)$$

Using the sum of an infinite geometric series $S=\frac{a}{1-r}$:
$$T_{\text{total}}=2\times \frac{0.4}{1-0.8}$$

Simplifying the fraction:
$$T_{\text{total}}=2\times \frac{0.4}{0.2}=2\times 2=4\text{seconds}$$

Therefore, the point mass comes to a complete rest on the ground after a total time of 4 seconds.