Two persons Ram and Shyam are throwing ball at each other as shown in the figure. The maximum horizontal distance from the building where Ram can stand and still throw a ball a Shyam is d1. The maximum horizontal distance of Ram from the building where Shyam can throw a ballis da. If both of them can throw ball with a velocity of 2gh, find out the ratio of d22d12.Neglect the height of each person.
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answer is 3.
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Detailed Solution
y=xtanθ−gx22u2sec2θWhen Ram throws the ball, h=xtanθ−gx22(2gh)21+tan2θtan2θ−8hxtanθ+1+8h2x2=0For real roots, D≥0⇒8hx2−4(1 +8h2x22≥0⇒ x≤8h⇒ xmax=8h d1=8hWhen Shyam throws the ball,y=xtanθ−gx22u2cos2θ⇒ −h=xtanθ−gx22(2gh)2sec2θ⇒tan2θ−8hxtanθ+1−8h2x2=0For real roots, D≥0 8hx2−41−8h2x2≥0 x≤24h xmax=24h d2=24h d22d12=24h8h2=3