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A particle just clears a wall of heig...

A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)b/(a c)` (2) `"45"^o` (3) `tan^(-1)(b c)/(a(c-a)` (4) `tan^(-1)(b c)/a`

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In the given diagram shown, for a projectile, what is the angle of projection? (A) tan^(-1) (1) (B) tan^(-1)(8/3) (C) tan^(-1)(4/3) (D) tan^(-1)(5/3)

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Knowledge Check

  • A particle just clear a wall of height b at a distance a and strikes the ground at a distance from the point of projection, the angle of projection is :

    A
    `tan^(-1)((b)/(ac))`
    B
    `45^(@)`
    C
    `tan^(-1)((bc)/(a(c-1)))`
    D
    `tan^(-1)((bc)/(a))`

  • In triangleABC , if angleA=90^(@)", then "tan^(-1)((c)/(a+b))+tan^(-1)((b)/(a+c))=

    A
    `(pi)/(8)`
    B
    `(pi)/(2)`
    C
    `(pi)/(4)`
    D
    `(pi)/(6)`

  • If angleA=90^@ in the triangleABC , then tan^(-1)(c/(a+b))+tan^(-1)(b/(a+c)) is

    A
    0
    B
    1
    C
    `pi/4`
    D
    `pi/6`

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    Let a, b and c be positive real numbers. Then prove that tan^(-1) sqrt((a(a + b + c))/(bc)) + tan^(-1) sqrt((b (a + b + c))/(ca)) + tan^(-1) sqrt((c(a + b+ c))/(ab)) = pi

    A vertical pole subtends an angle tan^(-1)((1)/(2)) at apoint P on the ground.The angle subtended by the upper half of the pole at the point P is (A)tan^(-1)((1)/(4))(B)tan^(-1)((2)/(9))(C)tan^(-1)((1)/(8))(D)tan^(-1)((2)/(3))

    A ball projected from ground at an angle of 45^(@) just clears a wall infront. If point of projection is 4m from the foot of wall and ball strikes the ground at a distance of 6m on the other side of the wall, the height of the wall is

    If in triangle ABC, C = 90^circ then prove that tan^(-1) (a/(c+b)) + tan^(-1) (b/(a+c)) = pi/4 .

    A ball is projected at an angle 45^(@) with horizontal. It passes through a wall of height h at horizontal distance d_(1) from the point of projection and strikes the ground at a horizontal distance (d_(1) + d_(2)) from the point of projection, then

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