A particle projected upwards with v = 20ms find time when the distance travelled is twice the displacemen?
Let it happen at a height of x meters.
Distance = twice |displacement|
=> H + (H-x) = 2 * x
=> 2H = 3x
=> x = 2/3 H
Now, v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time). We know v is zero at max height and u = 20m/s. Assuming acceleration due to gravity is 10m/s^2, we get
0 = 20 - 10t
=> t = 2
i.e. takes two seconds to reach max height.
Again, S = ut + 1/2at^2 (where S = distance). Substituting, we get
H = (20 * 2 ) - 1/2(10 * 2^2)
=> H = 20
thus x = 2/3H = 40/3
Now we just need to find out how long it takes for an object dropped from rest to fall a distance of H-x = 20 - 40/3 = 20/3
Using the equation S = ut + 1/2at^2, we get t = 2/root3
Thus, 2 seconds to get to max height and 2/root3 secs to fall, giving a total time requirement of 2 + 2/root3