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A body is projected at an angle 60^(@) w...

A body is projected at an angle `60^(@)` with the horizontal ground with kinetic energy k. When the velocity makes an angle `30^(@)` with the horizontal, the kinetic energy of the body will be `(k)/(a)`. Find value of a.

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To solve the problem, we will follow these steps: ### Step 1: Understand the initial conditions A body is projected at an angle of \(60^\circ\) with the horizontal with an initial kinetic energy \(K\). The initial velocity can be expressed as: \[ K = \frac{1}{2} m u^2 \] where \(u\) is the initial speed of the body and \(m\) is its mass. ### Step 2: Analyze the velocity components When the body reaches a point where the velocity makes an angle of \(30^\circ\) with the horizontal, we need to find the new velocity \(v\) at that point. The horizontal component of the velocity remains constant throughout the projectile motion. The horizontal component of the initial velocity is: \[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \] At the point where the angle is \(30^\circ\), the horizontal component of the velocity is: \[ v_x = v \cos(30^\circ) = v \cdot \frac{\sqrt{3}}{2} \] ### Step 3: Set the horizontal components equal Since the horizontal component of velocity remains constant, we can set the two expressions for horizontal velocity equal to each other: \[ \frac{u}{2} = v \cdot \frac{\sqrt{3}}{2} \] ### Step 4: Solve for \(v\) Rearranging the equation gives: \[ v = \frac{u}{\sqrt{3}} \] ### Step 5: Calculate the new kinetic energy The kinetic energy \(K'\) at the point where the velocity makes an angle of \(30^\circ\) is given by: \[ K' = \frac{1}{2} m v^2 \] Substituting \(v = \frac{u}{\sqrt{3}}\): \[ K' = \frac{1}{2} m \left(\frac{u}{\sqrt{3}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{3} = \frac{1}{3} \left(\frac{1}{2} m u^2\right) = \frac{K}{3} \] ### Step 6: Relate \(K'\) to \(K\) We know that \(K' = \frac{K}{a}\). From our previous result, we have: \[ \frac{K}{3} = \frac{K}{a} \] ### Step 7: Solve for \(a\) By comparing both sides, we find: \[ a = 3 \] ### Final Answer The value of \(a\) is \(3\). ---

To solve the problem, we will follow these steps: ### Step 1: Understand the initial conditions A body is projected at an angle of \(60^\circ\) with the horizontal with an initial kinetic energy \(K\). The initial velocity can be expressed as: \[ K = \frac{1}{2} m u^2 \] where \(u\) is the initial speed of the body and \(m\) is its mass. ...
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Knowledge Check

  • A body is projected at an angle 60^@ with horizontal with kinetic energy K. When the velocity makes an angle 30^@ with the horizontal, the kinetic energy of the body will be

    A
    `K//2`
    B
    `K//3`
    C
    `2K//3`
    D
    `3K//4`
  • If a body is projected with an angle theta to the horizontal, then

    A
    its velocity is always perpendicular to its acceleration
    B
    its velocity becomes zero at its maximum height.
    C
    its velocity makse zero angle with the horizontal at its maximum height
    D
    the body just before hitting the ground, the direction of velocity coincides with the acceleration
  • A ball is projected at an angle 60^(@) with the horizontal with speed 30 m/s. What will be the speed of the ball when it makes an angle 45^(@) with the horizontal ?

    A
    30 m/s
    B
    `15sqrt2m//s`
    C
    `(15)/(sqrt2)m//s`
    D
    `30sqrt2m//s`
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