Question

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

Answer 1

Hint- The height of the tower is h as given in the question. The ball takes T seconds to reach the ground. The initial velocity of the ball will be 0. The acceleration of the ball will be g. We will use the second equation of motion to solve this question.

Formula used: $s = ut + \dfrac{1}{2}a{t^2}$.

Complete step-by-step answer:

As the acceleration of the ball is $a = g$.
Initial velocity is $u = 0$.
And the time is $t = T$ seconds.
Distance travelled by the ball is equal to the height of the tower $s = h$.
Applying the second equation of motion i.e. $s = ut + \dfrac{1}{2}a{t^2}$ and putting the values of a, u, t and s, we get-
$
   \Rightarrow h = \left( 0 \right)T + \dfrac{1}{2}g{T^2} \\
    \\
   \Rightarrow h = \dfrac{1}{2}g{T^2} \\
$
Let the above equation be equation 1-
$ \Rightarrow h = \dfrac{1}{2}g{T^2}$ (equation 1)
Now, distance travelled in $\dfrac{T}{3}$ seconds-
$
   \Rightarrow {h_1} = \left( 0 \right)\dfrac{T}{3} + \dfrac{1}{2}g{\left( {\dfrac{T}{3}} \right)^2} \\
    \\
   \Rightarrow {h_1} = \dfrac{1}{2}g\dfrac{{{T^2}}}{9} \\
$
Let this above equation be equation 2-
$ \Rightarrow {h_1} = \dfrac{1}{2}g\dfrac{{{T^2}}}{9}$ (equation 2)
Dividing equation 2 by equation 1, we get-
$
   \Rightarrow \dfrac{{{h_1}}}{h} = \dfrac{{\dfrac{1}{2}g\dfrac{{{T^2}}}{9}}}{{\dfrac{1}{2}g{T^2}}} \\
    \\
   \Rightarrow \dfrac{{{h_1}}}{h} = \dfrac{1}{9} \\
    \\
   \Rightarrow {h_1} = \dfrac{h}{9} \\
$
Thus, $\dfrac{h}{9}$ is the distance from point of release.
Therefore, distance from the ground-
$
   \Rightarrow h - \dfrac{h}{9} \\
    \\
   \Rightarrow \dfrac{{8h}}{9} \\
$
Hence, option C is the correct option.

Note: Dividing equation 2 from equation 1 is necessary in order to find out the relation between given height and required height because options are present in terms of given height, So, do not forget to do so. The initial velocity is always 0 when the object moves from the stationary position.