Question

A ball is released from the top of a tower of height h meter. It takes T second to reach the ground. What is the position of the ball (from the ground) in T/3 second?
A. $\dfrac{h}{9}\,m$
B. $\dfrac{{7h}}{9}\,m$
C. $\dfrac{{8h}}{9}\,m$
D. $\dfrac{{17h}}{{18}}\,m$

Answer 1

As we all know that while dropping an object from the height, the initial velocity is zero. But in the case of throwing the object, the initial velocity is not zero. While dropping the ball from a height its velocity increases at each and every instant of time in a periodic manner.

Complete step by step solution:
Given: The height of the tower is $h\,$ metres from the top.
The time taken by the ball to reach the ground is $T$ seconds.
We can apply the equation of motion to exactly determine the relation of height with time till the ball reaches the ground.
As we all know that while dropping an object, equation of motion is given by,
$h = ut + \dfrac{1}{2}a{T^2}$…… (I)
Here $u$ is the initial velocity of the ball and $a$ is the acceleration.
We can see that in this case of dropping the values of $u$ and $a$ are:
$u = 0 \\\ a = g \\\$
Here $g$ is the acceleration due to gravity. Now we can substitute $u = 0$ and $a = g$ in equation (I). Therefore, we can notice that the equation (I) becomes,
$h = \dfrac{1}{2}g{T^2}$
This is the height a ball travels in time $T$.
Now let us suppose that the ball travels a distance equal to $x$ in the time $\dfrac{T}{3}$. So we can write equation of motion for this condition as:
$x = \dfrac{1}{2}g{\left( {\dfrac{T}{3}} \right)^2}$
$\Rightarrow x = \dfrac{1}{9}\left( {\dfrac{1}{2}g{T^2}} \right)$…… (II)
We can substitute $h = \dfrac{1}{2}g{T^2}$ in equation (II). So we can get,
$x = \dfrac{h}{9}$
Now this is the distance travelled by the ball from the tower in time $\dfrac{T}{3}$. So this distance from the ground is given by,
${x_1} = h - x \\\ \Rightarrow {x_1} = h - \dfrac{h}{9} \\\ \Rightarrow {x_1} = \dfrac{{8h}}{9} \\\$

$\therefore$ So the distance of the ball from the ground in time $\dfrac{T}{3}$ seconds is $\dfrac{{8h}}{9}$. The correct option is (C).

Note:
We know that while dropping the acceleration due to gravity $g$ is positive. We can also conclude that during any dropping and throwing ball problems the value of acceleration involved is $g$ and since we know that the distance varies with time in a periodic manner here as $h$ is proportional to ${T^2}$, so the distance traveled in each instant of time is more as compared to previous as the relation is not linear and while traveling the ball gains more and more velocity at each instant of time.