Three walls of height h, H, h are at equal distances a apart. A ball is thrown in a vertical plane perpendicular to
the wall so that it just clears all the walls. The ball strikes the ground at a distance x from the middle wall,
then x =

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Solution

The correct option is D
$H = \frac{A^{2}}{4 B}$ ; $2 x = \frac{A}{B} = R$
By Solving;
$A = \frac{2 H}{x}$ ; $B = \frac{H}{x^{2}}$
$h = A [x - a] - B [x - a]^{2}$ -----(i)
Substituting A and B value in eqn(i) we get; $h = \frac{2 H}{x} [x - a] - \frac{H}{x^{2}} [x - a]^{2}$
$x = a \sqrt{\frac{H}{H - h}}$ .

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Three walls of height h, H, h are at equal distances a apart. A ball is thrown in a vertical plane perpendicular to the wall so that it just clears all the walls. The ball strikes the ground at a distance x from the middle wall, then x =

The correct option is D H=A24B; 2x=AB=R By Solving; A=2Hx; B=Hx2 h=A[x−a]−B[x−a]2-----(i) Substituting A and B value in eqn(i) we get; h=2Hx[x−a]−Hx2[x−a]2 x=a√HH−h.

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the wall so that it just clears all the walls. The ball strikes the ground at a distance x from the middle wall,
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