Question

The distance described during the last half second of upward motion, of a body projected vertically upward 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: In order to find out the dependency of distance travelled by the body with other quantities like velocity of projection, mass or height then first of all use equation of motion for finding distance covered by the body using formula, ${\text{s = ut + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{a}}{{\text{t}}^{\text{2}}}$and then find out the distance covered by the body in last half second. Then use the equation of motion, ${\text{v = u + at}}$ and relate.

Complete step by step solution:
Let u be that initial velocity of the body
And acceleration due to gravity is g.
If the body covers maximum height in time n, then
Distance covered by the body in n seconds is given by
$\Rightarrow$ ${\text{s = un + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{n}}^{\text{2}}}$ $\left( {\because {\text{s = ut + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{a}}{{\text{t}}^{\text{2}}}} \right)$
Now, distance covered by the body in (n-1) second is given by
$\Rightarrow$ ${\text{s’ = u(n - 1) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g(n - 1}}{{\text{)}}^{\text{2}}}$
Distance travelled by the body in last second is given by
$
\Rightarrow {\text{D = s - s’}} \\
\Rightarrow {\text{D = un + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{n}}^{\text{2}}}{\text{ - u(n - 1) - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g(n - 1}}{{\text{)}}^{\text{2}}} \\
\Rightarrow {\text{D = un + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{n}}^{\text{2}}}{\text{ - un + u - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{n}}^{\text{2}}}{\text{ - g + 2gn}} \\
\Rightarrow \therefore {\text{ D = u - }}\dfrac{{\text{g}}}{{\text{2}}}{\text{ + gn}} \\
$
Now, distance travelled by the body in last half second is given by ${\text{n}} \to \dfrac{{\text{n}}}{{\text{2}}}$.
$\Rightarrow$ ${\text{D = u - }}\dfrac{{\text{g}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{g}}}{{\text{2}}}{\text{ = u}}...{\text{(i)}}$
For highest point using equation of motion, ${\text{v = u + gn}}$ ($\because {\text{v = u + at}}$)
$\Rightarrow$ ${\text{0 = u - 10n}}$
When ball is at highest point then the acceleration of gravity is in opposite direction to that of the ball so ${\text{g = 10 m/}}{{\text{s}}^{\text{2}}}$
$
\Rightarrow {\text{0 = u - 10n}} \\
   \Rightarrow {\text{u = 10n}} \\
$
Now substituting this value of initial velocity in (i), we get
$\Rightarrow$ ${\text{D = 10n}}$
So, when a body is projected vertically upward then the distance travelled by it in the last half second of its upward motion is independent of its initial velocity and therefore constant.

Therefore, option (C) is the correct choice.

Note: The total time taken by the object or particle to return to the same level from where it was thrown is known as time of flight and during the projection, the time of ascent is equal to the time of descent. A body thrown horizontally from a certain height above the ground follows a parabolic path till it hits the ground.