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

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).