$A$ particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momentum of the particle about $P$ at $t = \frac{V_0}{g}$ is:

  • A
    $\frac{1}{2 \sqrt{2}} \frac{mV_0^3}{g}$
  • B
    $\frac{1}{2 \sqrt{2}} \frac{mV_0^2}{g}$
  • C
    $\frac{1}{2} \frac{mV_0^3}{g}$
  • D
    $\frac{1}{2} \frac{mV_0^2}{g}$

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Similar Questions

$A$ body $P$ is projected at an angle of $30^{\circ}$ with the horizontal and another body $Q$ is projected at an angle of $30^{\circ}$ with the vertical. If the ratio of the horizontal ranges of the bodies $P$ and $Q$ is $1: 2$,then the ratio of the maximum heights reached by the bodies $P$ and $Q$ is

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$A$ body is moving with constant speed in a circle of radius $10 \ m$. The body completes one revolution in $4 \ s$. At the end of the $3^{rd}$ second,the displacement of the body (in $m$) from its starting point is:

$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

Fill in the blanks:
$(a)$ For vectors $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} + \hat{j} - 2\hat{k}$,the $y$-component of the vector $(\vec{A} - \vec{B})$ is .....
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