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Q.

A particle is projected from a point on ground. It just clears a vertical wall of height h at a distance h from the point of projection and finally hits the ground at a distance 3h from the base of the wall. Then maximum height attained by the projectile is

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a

$\frac{4 h}{3}$

b

2h

c

$\frac{3 h}{2}$

d

$\frac{5 h}{2}$

answer is A.

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Detailed Solution

Question Image

$y = x tanα (1 - \frac{x}{R})$

$∴ h = htanα (1 - \frac{h}{4 h})$

$\Rightarrow tanα = \frac{4}{3}$

$At x = \frac{R}{2} = 2 h$ , Height (y) attained by the projectile is maximum.

$∴ H = y_{\max} = 2 h . \frac{4}{3} (1 - \frac{2 h}{4 h}) = \frac{4 h}{3}$

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A particle is projected from a point on ground. It just clears a vertical wall of height h at a distance h from the point of projection and finally hits the ground at a distance 3h from the base of the wall. Then maximum height attained by the projectile is