If a stone is thrown vertically upward with a velocity 'u',the distance travelled during its last 't' seconds of motion is___

plz explain!

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)

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