Question

A body is thrown vertically up with a velocity $ u $ . It passes three points $ A,B $ and $ C $ in its upward journey with velocities $ \dfrac{u}{2} $ , $ \dfrac{u}{3} $ and $ \dfrac{u}{4} $ respectively. The ratio of the separation between the points $ A $ and $ B $ between $ B $ and $ C $ , that is, $ \dfrac{{AB}}{{BC}} $ is:
(A) $ 1 $
(B) $ 2 $
(C) $ \dfrac{{10}}{7} $
(D) $ \dfrac{{20}}{7} $

Answer 1

Hint:
Firstly, apply the formula given below and substitute the values at point $ A $ and again apply the same formula for point $ B $ and point $ C $ and find the separation between point $ A $ and $ B $ also find the separation between point $ B $ and $ C $ , finally we will put the values in $ \dfrac{{AB}}{{BC}} $ .
Here, we will use the formula given below:
 $\Rightarrow {v^2} - {u^2} = 2as $
Where, $ v $ = final velocity of the object, $ u $ = initial velocity of the object, $ a $ = acceleration of the object and $ s $ = displacement of the object.

Complete step by step solution:
Let the point $ A $ is located at the distance $ s $ from the ground
Now, the velocity at point $ A $ is given by:
 $\Rightarrow v = \dfrac{u}{2} $
Now using formula,
 $\Rightarrow {v^2} - {u^2} = 2as $
Now putting the value of $ v $ , we get
 $\Rightarrow \dfrac{{{u^2}}}{2} - {u^2} = 2( - g)s $
Here, the acceleration becomes acceleration due to gravity and we take it negative because we have thrown the body in upward direction.
Solving the above equation, we get
 $\Rightarrow \dfrac{{{u^2}}}{4} - {u^2} = - 2gs $
Again, solving the above equation, we get
 $\Rightarrow \dfrac{{ - 3{u^2}}}{4} = - 2gs $
Now, after solving we get the value of $ s $ , that is,
 $\Rightarrow s = \dfrac{{3{u^2}}}{{8g}}...(i) $
Now, the velocity at point $ B $ is given by:
 $\Rightarrow v = \dfrac{u}{3} $
Now using formula,
 $\Rightarrow {v^2} - {u^2} = 2as $
Now putting the value of $ v $ , we get
 $\Rightarrow {(\dfrac{u}{3})^2} = {u^2} + 2( - g){s^{'}} $
Solving the equation, we get
 $\Rightarrow {s^{'}} = \dfrac{{4{u^2}}}{{9g}}...(ii) $
Now, the velocity at point $ C $ is given by:
 $\Rightarrow v = \dfrac{u}{4} $
Now using formula,
 $\Rightarrow {v^2} - {u^2} = 2as $
Now putting the value of $ v $ , we get
 $\Rightarrow {(\dfrac{u}{4})^2} = {u^2} + 2( - g){s^{''}} $
Solving the above equation, we get
 $\Rightarrow {s^{''}} = \dfrac{{15{u^2}}}{{32g}}...(iii) $
Now, the separation between point $ A $ and $ B $ is given by:
 $\Rightarrow AB = {s^{'}} - s $
Putting the values from equation $ (i) $ and $ (ii) $ , we get
 $\Rightarrow AB $ = $ \dfrac{{4{u^2}}}{{9g}} $ - $ \dfrac{{3{u^2}}}{{8g}} $ =
 $\Rightarrow AB = \dfrac{{5{u^2}}}{{72g}} $
Now, the separation between $ B $ and $ C $ is given by,
 $\Rightarrow BC = {s^{''}} - {s^{'}} $
Putting the values, we get
 $\Rightarrow BC $ = $ \dfrac{{5{u^2}}}{{72g}} $ - $ \dfrac{{4{u^2}}}{{9g}} $
Solving the above equation, we get
 $\Rightarrow BC $ = $ \dfrac{{7{u^2}}}{{228g}} $
Now, the value of $ \dfrac{{AB}}{{BC}} $ is given by:
 $\Rightarrow \dfrac{{AB}}{{BC}} $ = $\dfrac{{\dfrac{{5{u^2}}}{{72g}}}}{{\dfrac{{7{u^2}}}{{228g}}}} $
 $\Rightarrow \dfrac{{AB}}{{BC}} $ = $ \dfrac{{5 \times 288}}{{72 \times 7}} $ = $ \dfrac{{20}}{7} $
Hence, the correct option is (D).

Note:
In the above solution, the acceleration due to gravity is taken as positive when the body is falling in downwards direction towards the earth surface and it is taken as negative when the body is thrown in the upward direction by some external force.