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Standard XII

Physics

Inertial vs Non Inertial Frames

A particle co...

Question

A particle covers $50 m$ distance when projected with an initial speed. On the same surface, when projected with double the initial speed it will cover a distance of :

100 m

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150 m

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200 m

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250 m

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Solution

The correct option is C 200 m
In projectile motion,
Range, $R = \frac{u^{2} s i n 2 θ}{g} = 50 m$ (Given)
Now, Initial speed is $u^{'} = 2 u$
$R^{'} = \frac{4 u^{2} s i n 2 θ}{g} = 4 \times 50$
$R^{'} = 200 m$
The correct option is C.

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The acceleration vector along x-axis of a particle having initial velocity $v_{0}$ changes with distance x as $a = \sqrt{x}$ .The distance covered by the particles, When its speed becomes twice that of initial speed is:-

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A particle covers 50m distance when projected with an initial speed. On the same surface, when projected with double the initial speed it will cover a distance of :

The correct option is C 200 mIn projectile motion,Range, R=u2sin2θg=50m (Given)Now, Initial speed is u′=2uR′=4u2sin2θg=4×50R′=200mThe correct option is C.

A particle covers $50 m$ distance when projected with an initial speed. On the same surface, when projected with double the initial speed it will cover a distance of :

The correct option is C 200 m
In projectile motion,
Range, $R = \frac{u^{2} s i n 2 θ}{g} = 50 m$ (Given)
Now, Initial speed is $u^{'} = 2 u$
$R^{'} = \frac{4 u^{2} s i n 2 θ}{g} = 4 \times 50$
$R^{'} = 200 m$
The correct option is C.