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(C) When the block is released from the horizontal position,it falls freely under gravity until the string becomes taut.The vertical distance fallen b

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$20$

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(C) When the block is released from the horizontal position,it falls freely under gravity until the string becomes taut.The vertical distance fallen by the block is $h = 5\, m$.The velocity $v_i$ of the block just before the string becomes tight is given by the equation of motion $v^2 = u^2 + 2gh$:$v_i = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 5} = 10\, m/s$ (directed vertically downwards).When the string becomes tight,the impulsive tension in the string acts upwards,bringing the vertical component of the velocity to zero instantaneously.Thus,the final vertical velocity $v_f = 0$.The impulse $I$ is equal to the change in momentum:$I = m(v_f - v_i) = 2(0 - 10) = -20\, N-s$.The magnitude of the impulse is $20\, N-s$.

$A$ block of mass $2\, kg$ is suspended by a string of length $5\, m$. It is released from a horizontal position (as shown in the figure) such that the string is initially slack. What will be the impulse when the string just becomes tight? (Take $g = 10\, m/s^2$)

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