Projection Under uniform Acceleration

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Question

A balll is projected with 20√2 m/s at angle 45⁰ with horizontal. The angular velocity of the particle at highest point of its journey about point of projection is

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Solution

$\large v = r\omega \Rightarrow \omega = \frac{v}{r}$
$\large \omega = \frac{{u\cos \theta \left( {h/r} \right)}}{r}$
$\large = \frac{{u\cos \theta h}}{{{r^2}}}$
$\large h = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ $\large =\frac{(20\sqrt2)^2(\frac{1}{\sqrt2})^2}{2x10}=20\;m$
$\large \frac{R}{2} = \frac{{{u^2}\sin 2\theta /g}}{2} = \frac{(20\sqrt2)^2(1)}{2x10}=40\;m$
r² = (40)² +(20)² = (20)² (5) m²
$\large \therefore \omega = \frac{20\sqrt2\;X\;\frac{1}{\sqrt2}X20}{(20)^2(5)} = \frac{1}{5} = 0.2{\text{ }}rad{\text{ }}{s^{ - 1}}$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A balll is projected with 20√2 m/s at angle 450 with horizontal. The angular velocity of the particle at highest point of its journey about point of projection is

r2 = (40)2 +(20)2 = (20)2 (5) m2

Ready to Test Your Skills?

free study material

A balll is projected with 20√2 m/s at angle 45⁰ with horizontal. The angular velocity of the particle at highest point of its journey about point of projection is