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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
  • a)
    100 m
  • b)
    150 m
  • c)
    200 m
  • d)
    250 m
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A particle covers 50 m distance when projected with an initial speed. ...
Explanation:
To solve this problem, we can use the formula for the range of a projectile, which is given by:

R = (v^2 sin 2θ) / g

where v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity. Since the surface is the same, we can assume that g is constant in both cases, and we can compare the two ranges using the ratio of their initial speeds.

Let's assume that the initial speed in the first case is v, and the corresponding angle of projection is θ. Then we have:

50 = (v^2 sin 2θ) / g

Now, let's assume that the initial speed in the second case is 2v, and the corresponding angle of projection is also θ. Then we have:

R = [(2v)^2 sin 2θ] / g

R = (4v^2 sin 2θ) / g

Therefore, the ratio of the two ranges is:

R / 50 = [(4v^2 sin 2θ) / g] / [v^2 sin 2θ / g]

R / 50 = 4

R = 200 m

Therefore, the particle will cover a distance of 200 m when projected with double the initial speed. The correct answer is option C.
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Community Answer
A particle covers 50 m distance when projected with an initial speed. ...
In a projectile motion, R=u^2sin2 theta/g=50 (eq 1)

in 2nd case
initial speed is u'.
so,R'=(2u)^2sin2 theta/g
=4u^2sin2 theta/g
=4×50. (from eq 1)
=200m


so the correct answer is 200m.
option c
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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 speeda)100 mb)150 mc)200 md)250 mCorrect answer is option 'C'. Can you explain this answer?
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