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Question
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 of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is 19.6 cm from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationery goli without falling on the ground earlier?
Solution
Given:
Distance between the golis of the first and second players = 2.0 m = R = Horizontal range
Height h from which the goli is projected by the second player = 19.6 cm = 0.196 m
We know that the goli moves in projectile motion.
Acceleration due to gravity a = g = 9.8 m/s2
The time in which the goli will reach the ground is given by the equation of motion.
As the initial velocity uy in vertical direction is zero, we have:
\[s = u_y t + \frac{1}{2}g t^2\]
\[t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0 . 196}{9 . 8}}\]
\[ = 0 . 4 = 0 . 2 s\]
Let us assume that the goli is projected with horizontal velocity ux m/s.
The horizontal range is given by
R = uxt
\[\Rightarrow u = \frac{R}{t} = \frac{2}{0 . 2} = 10 \text{ m }/s\]
Hence, if the second player projects the goli with a speed of 10 m/s, then his goli will hit the goli of the first player.
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