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For a given velocity, a projectile has t...

For a given velocity, a projectile has the same range `R` for two angles of rpojection if `t_(1)` and `t_(2)` are the times of flight in the two cases then

A

`t_1 t_2 prop R`

B

`t_1 t_2 prop R^2`

C

`t_2t_2 prop 1/(R^2)`

D

`t_1t_2 prop 1/R`

Text Solution

Verified by Experts

The correct Answer is:
A
`t_1 = (2u sin alpha)/(g)`
`t_2 = (2 u sin (90^@ - alpha))/(g)`
So `t_1 xx t_2 = 2 (u^2)/(g^2) sin 2 alpha`
or `t_1 xx t_2 = (2R)/(g) " " (because R = (u^2 sin alpha)/(g))`
`t_1 xx t_2 prop R`
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Knowledge Check

  • For a given velocity, a projectile has the same range R for two angles of projection. If t_(1) and t_(2) are the time of flight in the two cases, then t_(1) = t_(2) is equal to

    A
    `(2R)/(g)`
    B
    `(R)/(g)`
    C
    `(4R)/(g)`
    D
    `(R)/(2g)`

  • A projectile can have the same range R for two angles of projection. If t_(1) and t_(2) be the times of flight in the two cases:-

    A
    `t_(1)t_(2)propR^(2)`
    B
    `t_(1)t_(2)propR`
    C
    `t_(1)t_(2)prop(1)/(R)`
    D
    `t_(1)t_(2)prop(1)/(R_(2))`

  • A projectile has the same range R for two angles of projections but same speed. If T_(1) and T_(2) be the times of flight in the two cases, then Here theta is the angle of projection corresponding to T_(1) .

    A
    `T_(1)T_(2) prop R`
    B
    `T_(1)T_(2)propR^(2)`
    C
    `(T_(1))/(T_(2))=tantheta`
    D
    `(T_(1))/(T_(2))=cot theta`

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