Question

A stone is dropped from the top of top of a height h. After 1s another stone is dropped from the balcony 20m below the top. Both reach the bottom simultaneously. What is the value of h? Take $g = 10m{s^{ - 2}}$.
$
  (a){\text{ 3125m}} \\
  (b){\text{ 312}}{\text{.5m}} \\
  (c){\text{ 31}}{\text{.25m}} \\
  (d){\text{ 25}}{\text{.31m}} \\
$

Answer 1

Hint – In this question use the concept that when a stone is dropped from top of a tower the initial velocity of the stone is 0. So let the stone take t sec to reach the bottom, apply the second law of motion that is $S = ut + \dfrac{1}{2}a{t^2}$. The height from which the second stone is dropped is (h-20 m) and t is (t-1 sec) again the application of the second law of motion will help formation of quadratic equations in time. This will help approaching the problem.
Step by step answer:
If the stone is dropped from the top of the tower then stone initial velocity should be zero.
Let it drop from a height h meter, so the height of the tower is h meter.
And let it take t sec to reach the ground.
So in t sec stone covers h meter distance.
Now according to second law of motion we have,
$ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}$..................... (1)
Where s = distance covered in t sec having (u) initial velocity and (a) acceleration.
In case of stone s = h meter, u = 0m/s t = t sec and a = g (acceleration due to gravity = 10m/s2).
$ \Rightarrow h = \left( 0 \right)t + \dfrac{1}{2}g{t^2}$
$ \Rightarrow h = \dfrac{1}{2}\left( {10} \right){t^2} = 5{t^2}$meters................ (2)
Now a second stone drops after 1 sec from the balcony 20 m below the top and it reaches the bottom at the same time.
So the height of the balcony from the ground = (h – 20) m.
And it takes (t – 1) sec to reach the ground.
And the initial velocity of the second stone is also zero.
Therefore for second stone, s = h – 20, u = 0 m/s, t = (t – 1) sec and a = g = 10m/s2
Now from equation (1) we have,
$ \Rightarrow h - 20 = \left( 0 \right)t + \dfrac{1}{2}\left( {10} \right){\left( {t - 1} \right)^2}$
$ \Rightarrow h - 20 = 5\left( {{t^2} + 1 - 2t} \right)$............... (3), $\left[ {\because {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right]$
Now from equation (2) substitute the value of h in equation (3) we have,
$ \Rightarrow 5{t^2} - 20 = 5{t^2} + 5 - 10t$
Now simplify this we have,
$ \Rightarrow - 20 = + 5 - 10t$
$ \Rightarrow 10t = 25$
$ \Rightarrow t = \dfrac{{25}}{{10}} = 2.5$ Seconds.
Now from equation (2) we have,
$ \Rightarrow h = 5{\left( {2.5} \right)^2} = 5\left( {6.25} \right) = 31.25$m.
So this is the required height of the tower.
Hence option (c) is the correct answer.

Note – The trick point here was that the laws of equation of motion are only applicable if the acceleration is constant as for a ball falling down the acceleration due to gravity is constant that is g in the downwards direction . g is not taken as negative because downwards direction is taken as positive. Other laws of the equation of motion states $v = u + at,{v^2} - {u^2} = 2as$.