A particle is projected vertically upwards with a velocity of 20m/sec. Find the time at which distance travelled is twice the displacement

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$2 + \sqrt{4 / 3} \sec$

1 sec

$2 + \sqrt{3 / 4} \sec$

3sec

answer is A.

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Detailed Solution

From the figure below, at point C, the distance traveled is twice the displacement.

For AB, using the third equation of motion we have,

v²− u²= 2(g)(H_max)

⇒ 0 - (20)(20) = −2(10).(3s/2)

⇒ s = (20×20)/(10×3) = 40/3

Now, using the second equation of motion we have,

s = ut−1/2gt²

⇒ 40/3 = 20t − 1/2×10t²

3t²− 12t+8=0

On solving t = 2 +√(4/3).

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