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A cricketer hits a ball in a vertical x-...

A cricketer hits a ball in a vertical x-y plane from the ground level with a velocity `barV_0=(10sqrt(3i)+30hatj)` m/s.Find the time in which velocity vector makes an angle of `30 ^(@)` with horizontal x- axis (g=10 `m//s^(2))`

Text Solution

AI Generated Solution

To solve the problem step by step, we will analyze the motion of the ball hit by the cricketer and determine the time at which the velocity vector makes a 30-degree angle with the horizontal x-axis. ### Step 1: Identify the Components of the Initial Velocity The initial velocity vector of the ball is given as: \[ \vec{V_0} = 10\sqrt{3} \hat{i} + 30 \hat{j} \text{ m/s} \] Here, the x-component \(V_{0x} = 10\sqrt{3}\) m/s and the y-component \(V_{0y} = 30\) m/s. ...
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Knowledge Check

  • A ball is projected with a velocity 20 sqrt(3) ms^(-1) at angle 60^(@) to the horizontal. The time interval after which the velocity vector will make an angle 30^(@) to the horizontal is (Take, g = 10 ms^(-2))

    A
    5s
    B
    2s
    C
    1s
    D
    3s

  • A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

    A
    10m
    B
    5m
    C
    2.5m
    D
    1m

  • A particle is projected from the ground at an angle of 60^(@) with horizontal at speed u = 20 m//s. The radius of curvature of the path of the particle, when its velocity. makes an angle of 30^(@) with horizontal is : (g=10 m//s^(2)

    A
    `10.6 m`
    B
    `12.8 m`
    C
    `15.4 m`
    D
    `24.2 m`

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