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The equation of trajectory of an oblique...

The equation of trajectory of an oblique projectile `y = sqrt(3) x - (g x^(2))/(2)`. The angle of projection is

A

`30^(@)`

B

`45^(@)`

C

`60^(@)`

D

`15^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`y = sqrt(3) x - (g x^(2))/(2)`
The equation of trajectory of projectile
`y = x tan theta - (g x^(2))/(2u^(2) cos^(2) theta)`
Comparing coefficient of `x`
`tan theta = sqrt(3) rArr theta = 60^(@)`
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Knowledge Check

  • The equation of a projectile is y=sqrt(3)x-(gx^(2))/(2) the angle of projection is:-

    A
    `30^(@)`
    B
    `60^(@)`
    C
    `45^(@)`
    D
    none

  • Equation of trajector of ground to ground projectile is y=2x-9x^(2) . Then the angle of projection with horizontal and speed of projection is : (g=10m//s^(2))

    A
    `tan^(-1)(2),(5)/(2)m//s`
    B
    `tan^(-1)(3),(5)/(3)m//s`
    C
    `tan^(-1)(2),(2)/(3),(2)/(3)m//s`
    D
    `tan^(-1)(3),(2)/(3)m//s`

  • An object is projected with a velocity of 10 m//s at an angle 45^(@) with horizontal. The equation of trajectory followed by the projectile is y = a x - beta x^(2) , the ratio alpha//beta is

    A
    `5`
    B
    `10`
    C
    `15`
    D
    `20`

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