Question bank/JEE Main/Physics

Two stones are in the same horizontal line and are$50\sqrt{5}m$
away. They are projected with a velocity of 10
respectively and with an angle of 37º and 53º with horizontal
respectively. The minimum distance between them in meters is 5n. Find s.
Suppose they don't strike the ground during the motion.
(take$g = 10m/s^{2}$)

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$d_{\min} = 50\sqrt{5}\ \ \ \ \ \ \ \ sin\theta = 50\sqrt{5} \times \frac{2}{5\sqrt{5}} = 20m$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx