An object of mass 'm' is projected from origin in a vertical xy-plane at an angle 45° with the x-axis with an initial velocity . The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum heigh

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Solution

Q.
An object of mass 'm' is projected from origin in a vertical xy-plane at an angle 45° with the x-axis with an initial velocity $v_{0}$ . The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity] [2025]

1 $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ along negative z-axis

2 $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ along positive z-axis

3 $\frac{m v_{0}^{3}}{4 \sqrt{2} g}$ along positive z-axis

4 $\frac{m v_{0}^{3}}{4 \sqrt{2} g}$ along negative z-axis

Ans.
(4)

At highest point

The speed is $v_{x} = v_{0} \cos 45 ° \Rightarrow v_{x} = \frac{v_{0}}{\sqrt{2}}$

$H = \frac{{(\frac{v_{0}}{\sqrt{2}})}^{2}}{2 g} = \frac{v_{0}^{2}}{4 g}$

So, angular momentum about origin is

$L = m v_{x} \cdot H$

$\Rightarrow L = m \times \frac{v_{0}}{\sqrt{2}} \times \frac{v_{0}^{2}}{4 g}$

$\Rightarrow L = \frac{m v_{0}^{3}}{4 \sqrt{2} g}$ (along negative z-axis)

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