If a stone is thrown vertically upward with a velocity 'u',the distance travelled during its last 't' seconds of motion is___
Let t be the time taken to reach the maximum height.
So, using, v = u + at
=> 0 = u – gt
=> t = u/g
Now, time of ascent = time of decent
So, while falling down, initial velocity is zero.
So, distance travelled in time t, just before hitting the ground is, St = 0 + ½ gt2
Distance travelled in time (t-1) s is, St-1 = 0 + ½ g(t-1)2
So, the distance travelled in the last second is,Slast = St – St-1 = ½ gt2 - ½ g(t-1)2
=> Slast = ½ g(t2 – t2 +2t -1)
=> Slast = ½ g(2t - 1)