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.

  • $\begingroup$ Why is the initial velocity 9.8m/s in the upward direction? $\endgroup$ – tatan Sep 5 '15 at 17:13
  • $\begingroup$ It's not the initial velocity. It is the velocity at beginning of the last second it travels before it comes to rest at the top. It's 9.8 m/s because g=9.8m/s^2 acts on it for one second and makes it 0. $\endgroup$ – sarat.kant Sep 5 '15 at 17:15
  • $\begingroup$ The velocity changes at 9.8m/s per second in the downward direction (as acceleration is in downward direction).Why are you saying then the direction of motion is in upward direction with a 9.8m/s? $\endgroup$ – tatan Sep 5 '15 at 17:18
  • $\begingroup$ @tatan okay here's another way to look at it. Since the upward motion and the downward motion are completely identical in terms of distance travelled per second. The distance travelled by the object in it's last second of upward motion is the same as the distance travelled in it's first second of downward motion. Since it starts from rest in it's downward motion, this distance is constant and independent of anything else. $\endgroup$ – sarat.kant Sep 5 '15 at 17:28
  • $\begingroup$ Nice explanation!! $\endgroup$ – tatan Sep 5 '15 at 17:36

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