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[" If direction of velocity of a projectile makes an angle "],[" of "37^(@)" with the horizontal,at a time which is "1" s before "],[" the time of reaching the maximum height,then velocity "],[" at highest point,will be "(g=10m/s^(2))]

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A projectile is fired at an angle theta with the horizontal with velocity u. Deduce the expression for the maximum height reached by it.

A particle is projected with a velocity of 25 m//s at an angle of 30^(@) with the horizontal . Calculate (i) The maximum height , (ii) time of flight and (iii) horizontal range .

Knowledge Check

  • After one second the velocity of a projectile makes an angle of 45^(@) with the horizontal. After another one more second its is travelling horizontally. The magnitude of its initial velocity and angle of projectile are (g=10m//s^(2))

    A
    `14.62m//s, 60^(@)`
    B
    `14.62m//s, tan^(-1)(2)`
    C
    `22.36m//s, tan^(-1)(2)`
    D
    `22.36 m//s, 60^(@)`

  • Velocity of a stone projected, 2 second before it reaches he maximum height, makes angle 53^(@) with the horizontal then the velocity at highest point will be

    A
    `20 m//s`
    B
    `15 m//s`
    C
    `25 m//s`
    D
    `80//3 m//s`

  • Velocity of a stone projected, 2 second bofore it reaches the maximum height makes angle 53^@ with the horizontal then the velocity at highest point will be

    A
    `20(m)/(s)`
    B
    `15(m)/(s)`
    C
    `25(m)/(s)`
    D
    `(80)/(2)(m)/(s)`

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    A projectile is thrown with velocity v at an angle theta with the horizontal. When the projectile is at a height equal to half of the maximum height,. The vertical component of the velocity of projectile is.

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