Question

A body is thrown vertically upward from a point ‘A’ $125\,{\text{m}}$ above the ground. It goes up to maximum height of $250\,{\text{m}}$ above the ground and passes through ‘A’ on its downward journey. The velocity of the body when it is at a height of $70\,{\text{m}}$ above the ground is: $\left( {g = 10\,{\text{m}}{{\text{s}}^{ - 2}}} \right)$
(A) $60\,{\text{m}}{{\text{s}}^{ - 1}}$
(B) $80\,{\text{m}}{{\text{s}}^{ - 1}}$
(C) $20\,{\text{m}}{{\text{s}}^{ - 1}}$
(D) $50\,{\text{m}}{{\text{s}}^{ - 1}}$

Answer 1

This is the case of free fall from the maximum height. Use the equation of motion, put the values and manipulate accordingly to find the result. Don’t take $70\,{\text{m}}$ as the distance traversed by the body.

Complete step by step answer:
-Given, the body is thrown upward with a certain velocity, for which the body reaches a maximum height of $250\,{\text{m}}$ above the ground. In this case, the initial velocity is the velocity with which the body is thrown upward and the final velocity of the body becomes zero on reaching the maximum height. This height is that distance above the ground, beyond which the body won’t move any further.
-On reaching the maximum height, the body now will move downward. During the downward motion, the final velocity of the upward motion will become the initial velocity for the downward motion. We are asked to calculate the velocity of the body when it is at a height of $70\,{\text{m}}$ above the ground. So, during the downward motion:
Initial velocity $\left( u \right)$ of the body is $60\,{\text{m}}{{\text{s}}^{ - 1}}$
Distance $\left( S \right)$ traversed by the body is $\left( {250 - 70} \right)\,{\text{m}}$
To find the velocity at the mentioned height, we use the formula of motion:
${v^2} = {u^2} + 2gS$ …… (1)
Where,
$v$ indicates final velocity.
$u$ indicates initial velocity.
$g$ indicates the acceleration due to gravity.
$S$ indicates the distance traversed.
Substituting the values in the equation (1), we get:

$${v^2} = {u^2} + 2gS \\\ \Rightarrow{v^2} = {0^2} + 2 \times 10 \times \left( {250 - 70} \right) \\\ \Rightarrow{v^2} = 2 \times 10 \times 180 \\\$$

Simplifying the above equation, we get:

$${v^2} = 3600 \\\ \Rightarrow v = \sqrt {3600} \\\ \therefore v = 60\,{\text{m}}{{\text{s}}^{ - 1}} \\\$$

Hence, the velocity of the body when it is at a height of $70\,{\text{m}}$ above the ground is $60\,{\text{m}}{{\text{s}}^{ - 1}}$. The correct option is (A).

Note: While solving this problem, you must be well versed in the concept of initial and final velocity. Remember that, final velocity for the upward motion becomes the initial velocity for the downward motion. The distance traversed by the body is not $70\,{\text{m}}$, rather it is the position at which we are asked to calculate the velocity.