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During a projectile motion, if the maxim...

During a projectile motion, if the maximum height equals the horizontal range, then the angle of projection with the horizontal is.

A

`45^(@)`

B

`theta = tan^(-1)(0.25)`

C

`theta = tan^(-1)4`

D

`60^(@)`

Text Solution

Verified by Experts

The correct Answer is:
c
`R = H, v_(0)^(2) sin2 theta//g = v_(0)^(2) sn^(2) theta//2g, theta = tan^(-1) (4)`
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Knowledge Check

  • If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

    A
    `tan^(-1)((H)/(R ))`
    B
    `tan^(-1)((2H)/(R ))`
    C
    `tan^(-1) ((4H)/(R ))`
    D
    `tan^(-1) ((4H)/(H))`

  • The horizontal range of projectile is 4sqrt(3) times the maximum height achieved by it, then the angle of projection is

    A
    `30^(@)`
    B
    `45^(@)`
    C
    `60^(@)`
    D
    `90^(@)`

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