Question

The distance described during the last half second of upward motion, of a body projected vertically upwards is:
(A) Dependent upon the velocity of projection
(B) Dependent upon the time taken to reach maximum height
(C) Always constant
(D) Dependent upon mass of the body

Answer 1

Hint
To find the distance travelled by a body projected upwards in the last second, we will first find the total distance travelled by the body in time ‘t’ to attain the maximum height. Then we will find the distance travelled by the body in ‘ $t - 1$ ’ time. Using these two equations we will get the distance travelled by the body in the last second. This will give us an idea about the dependence of distance on various factors.

Complete step by step answer
We will first determine the distance total distance travelled to reach the maximum height.
$\Rightarrow {S_t} = ut - \dfrac{1}{2}g{t^2}$ …………………..(i)
(where u is the speed of projection and t is the total time taken to reach the maximum height)
Now, the distance travelled in the last second.
$\Rightarrow {S_{t - 1}} = u(t - 1) - \dfrac{1}{2}g{\left( {t - 1} \right)^2}$ ……………………..(ii)
On Subtracting eq.(ii) from eq.(i), we get
$\Rightarrow {S_t} - {S_{t - 1}} = u - gt + \dfrac{g}{2}$ ……………….(iii)
Eq.(iii) is the distance travelled by the body in the last second.
We know that, when a body is projected vertically upwards, $u = gt$ (this is because ‘t’ is the time when it reaches the maximum height)
Therefore, on substituting the value of u in eq.(iii), we get
$\Rightarrow {S_t} - {S_{t - 1}} = gt - gt + \dfrac{g}{2}$
$\Rightarrow {S_t} - {S_{t - 1}} = \dfrac{g}{2}$
The above relation tells us that the distance travelled by a body projected vertically upwards in the last second is a constant quantity, independent of speed, time travelled and mass of the body.
Hence, the distance described during the last half second of upward motion, of a body projected vertically upwards is always constant.
The correct answer is option (C).

Note
When the motion takes place under the effect of gravitational attractive force only, the motion is known as free fall. While solving a question for free fall, make sure you follow the correct sign convention.
- Displacement will be taken as positive if the final position lies above initial position and negative if the position lies below initial position.
- Velocity will be taken as positive if it is upward and negative if it is downward.
- Acceleration ‘a’ is always taken to be ‘-g’.