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A football is kicked at an angle of 30^(...

A football is kicked at an angle of `30^(@)` with the vertical, so if the horizontal component of its velocity is `20 ms^(-1)`, determine its maximum height.

Text Solution

Verified by Experts

Given, `theta = 30^(@)`
Horizontal component of velocity `= u sin 30^(@) = 20 ms^(-1)`
`u = (20)/(sin 30^(@)) = (20)/(1//2) = 40 ms^(-1)`
Therefore, maximum height,
`H = (u^(2) cos^(2) 30^(@))/(2g) = ((40)^(2))/(2 xx 9.8) xx ((sqrt(3))/(2))^(2) = 61.22 m`
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Knowledge Check

  • Assume that a ball is kicked at an angle of 60^(@) with the horizontal, so if the horizontal component of its velocity is 19.6 ms^(-1) , determine its maximum height.

    A
    `58.8 m`
    B
    `40 m`
    C
    `120 m`
    D
    `60 m`
  • The velocity of a body is 20 m/s making an angle of 30^(@) with the horizontal, the vertical component of velocity is

    A
    20m/s
    B
    17.32m/s
    C
    10m/s
    D
    7m/s
  • A bu llet is fired with a velocity u making an angle of 60^(@) with the horizontal plane. The horizontal component of the velocity of the bu llet when it reaches the maximum height is:

    A
    `u `
    B
    0
    C
    `(sqrt(3)u)/(2)`
    D
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