A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb

The key to solving this problem is to recognize that the horizontal velocity of the projectile is constant and there is no horizontal acceleration. Therefore, the time taken by the projectile to travel {eq}2.0 m {/eq} horizontally is given by: {eq}time = \frac{distance}{velocity} {/eq} We need to find the velocity required to hit the stationary goli which is {eq}2.0 m {/eq} away. We can split this problem into two parts - finding the height of the projectile at the point of impact and finding the initial velocity required to reach that height. First, we need to find the time taken by the projectile to travel horizontally {eq}x=2.0 m {/eq}. We know that the acceleration due to gravity is {eq}g=9.8 m/s^2{/eq} and the initial vertical velocity is zero. Using the following kinematic equation: {eq}x = v_{x} t {/eq} where {eq}v_{x} {/eq} is the horizontal velocity, we can calculate the time of flight {eq}t {/eq}: {eq}t = \frac{x}{v_{x}} {/eq} Next, we need to find the height {eq}h {/eq} of the projectile at the point of impact. We can use the following kinematic equation: {eq}h = v_{i}t - \frac{1}{2}gt^2 {/eq} where {eq}v_{i} {/eq} is the initial vertical velocity. At the point of impact, the height {eq}h {/eq} is zero. Therefore, we can write: {eq}0 = v_{i}t - \frac{1}{2}gt^2 {/eq} Solving for {eq}v_{i} {/eq}, we get: {eq}v_{i} = \frac{1}{2}gt {/eq} Substituting the value of {eq}t {/eq} from the first equation, we get: {eq}v_{i} = \frac{x}{t} times \frac{1}{2}g {/eq} Substituting the given values, we get: {eq}v_{i} = \frac{2.0 m}{0.407 s} times \frac{1}{2}(9.8 m/s^2) \approx 9.6 m/s {/eq} Therefore, the goli should be projected at a speed of about {eq}9.6 m/s {/eq} horizontally to hit the stationary goli without falling on the ground earlier.