I recently read in a book that the distance covered by an object projected vertically upwards in the last second of the upward motion is a constant independent of initial velocity.I tried to prove it using the formula -$S_n=u+\frac 12a(2n-1)$ but could not proceed from there.Please help to do it.thanks a lot in advance.
The last second of it's upward motion means that, at the end of that second, the object is at rest. It will start falling back downwards after that. Now since we know that the body experiences constant deceleration $g$ (= 9.8 $m/s^2$), it will have a velocity $u$ of 9.8 $m/s$ in the upward direction at the beginning of that second. Now since we have the initial velocity, time travelled and acceleration of the body, distance travelled can be calculated using $$ S = ut+ 1/2at^2 $$ Where $u = 9.8m/s$, $t = 1sec$, $a = -9.8 m/s^2$. This gives a value of 4.9m for $S$ which is constant and not dependent on the initial velocity of the object.