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

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

$R = \frac{u^{2} s i n 2 θ}{g} = R \propto u^{2}$ .
o if the speed of projection doubled, the range will becomes four times, i.e., $4 \times 50 = 200 m$

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A

is projected with an initial velocity of

60 m / s

at an angle

30^{\circ}

to the horizontal. At the same time, a second particle

B

is projected in direction as shown in figure with initial speed of

50 m / s

from a point at a distance of

100 m

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A

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

The correct option is C 200 m R=u2sin2θg=R∝u2. o if the speed of projection doubled, the range will becomes four times, i.e., 4×50=200 m

The correct option is C
200 m

$R = \frac{u^{2} s i n 2 θ}{g} = R \propto u^{2}$ .
o if the speed of projection doubled, the range will becomes four times, i.e., $4 \times 50 = 200 m$