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The speed of a projectile (u) rekuces by...

The speed of a projectile (u) rekuces by ` 50 %` on reachig maximum hight. What is the range on the horizontal plane ?

Text Solution

Verified by Experts

If ` theta` is the angle of projection, then velocity of projectile at height point `= u cos theta`
As per question, ` u cos theta = ( 50)/( 100) u = 1/ 2 u ` or ` cos theta = 1/2 = cos 60^@` or ` theta =60^@`
Horizontal range, `R= ( u^2 sin 2 theta)/g = u^2 /g sin 2 xx 60^@ = u^2 /g sin 120^@ = u^2/g xx sin ( 180^@- 60^@)`
` u^2/g sin 60^@ = u^@/g 60^2 = u^2/g xx (sqrt 3)/2`.
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Knowledge Check

  • A body is thrown up with a speed u , at an angle of projection theta If the speed of the projectile becomes u/sqrt2 on reaching the maximum height , then the maximum vertical height attained by the projectile is

    A
    `(u^2)/(4g)`
    B
    `(u^2)/(3g)`
    C
    `u^2/(2g)`
    D
    `(u^2)/g`

  • The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is

    A
    `(3 u^(2))/(g)`
    B
    `(sqrt(3) u^(2))/(2 g)`
    C
    `(3u^(2))/(g^(2))`
    D
    `(u^(2))/(3g)`

  • The speed at the maximum height of a projectile is sqrt(3)/(2) times of its initial speed 'u' of projection Its range on the horizontal plane:-

    A
    `sqrt(3)(u^(2))/(2g)`
    B
    `(u^(2))/(2g)`
    C
    `(3u^(2))/(2g)`
    D
    `(3u^(2))/(g)`