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For angles of projection of a projectile...

For angles of projection of a projectile at angle `(45^(@) - theta) and (45^(@)+ theta)`, the horizontal ranges described by the projectile are in the ratio of :

A

`1 :1 `

B

` 2 : 3`

C

`1 : 2`

D

` 3 : 2`

Text Solution

Verified by Experts

The correct Answer is:
A
Angle `(45^(@)+theta)` , Range = RI
`(45^(@)+theta)` , Range = `R_(2)`
`(R_(1))/(R_(2))=([(u^(2) sin 2 (45^(@)= theta))/(g)])/([[u^(2) sin 2(45^(@) + theta))/(g)])=(u^(2) sin(90-20))/(u^(2) sin (90 +20))`
`(cos 2 theta)/(cos 2 theta)=1`
`:. R_(1)=R_(2)` Hence, for complementary angles, ranges will be same .
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Knowledge Check

  • Galileo writes that for angles of projection of a projectile at angles (45 + theta) and (45 - theta) , the horizontal ranges described by the projectile are in the ratio of (if theta le 45 )

    A
    `2 : 1`
    B
    `1: 2`
    C
    `1 : 1`
    D
    `2 : 3`

  • For angles of projection of a projectile at angles (45^(@) – θ) and (45^(@) + θ) , the horizontal ranges described by the projectile are in the ratio of:

    A
    ` 1 : 1 `
    B
    `2 : 3 `
    C
    ` 1 : 2 `
    D
    `2 : 1 `

  • For angle of projection of a projectile at angles and (45^(@)+theta) , the horizontal range described by the projectile are in the ratio of

    A
    `2:1`
    B
    `1:1`
    C
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    D
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