Question

Two balls are dropped from the same height from places A and B. The body at B takes two seconds less to reach the ground at B strikes the ground with a velocity greater than at A by $10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$. The product of the acceleration due to gravity at the two places A and B is:
A. 5
B. 25
C. 125
D. 12.5

Answer 1

We will write the expression for ball velocity and time taken by the balls to reach the ground at place A and place B, respectively. We will substitute their values in the mathematical equations of the given statement of the problem.

Complete step by step answer:
Assume:
The velocity of the ball at place A is ${v_A}$.
The velocity of the ball at place B is ${v_B}$.
The time taken by the ball at place A to reach the ground is ${t_A}$.
The time taken by the ball at place B to reach the ground is ${t_B}$.
The acceleration due to gravity at place A is ${g_A}$.
The acceleration due to gravity at place B is ${g_B}$.
We have to find the product of acceleration due to gravity at places A and B.
It is given that the ball at place B takes two seconds less than the ball at place A to reach the ground so we can write:
${t_A} - {t_B} = 2{\rm{ s}}$……(1)
It is also given that the velocity of the ball at place B is greater than the ball at place A by $10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$ so we can write:

{v_B} = {v_A} + 10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}\\\ {v_B} - {v_A} = 10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{array}$$……(2) On dividing equation (1) and equation (2), we get: $$\dfrac{{{v_B} - {v_A}}}{{{t_A} - {t_B}}} = \dfrac{{10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{2{\rm{ s}}}}$$……(3) Let us write the expression for the ball's velocity at place A and place B, respectively. $${v_A} = \sqrt {2{g_A}h} $$ And, $${v_B} = \sqrt {2{g_B}h} $$ Here h is the distance between the ground and place A and B. We can write the expression for the time taken by ball at place A and place B to reach the ground as below: $${t_A} = \sqrt {\dfrac{{2h}}{{{g_A}}}} $$ And, $${t_B} = \sqrt {\dfrac{{2h}}{{{g_B}}}} $$ Substitute $$\sqrt {2{g_A}h} $$ for $${v_A}$$, $$\sqrt {2{g_B}h} $$ for $${v_B}$$, $$\sqrt {\dfrac{{2h}}{{{g_A}}}} $$ for $${t_A}$$, and $$\sqrt {\dfrac{{2h}}{{{g_B}}}} $$ for $${t_B}$$ in equation (3). $$\begin{array}{l} \dfrac{{\sqrt {2{g_B}h} - \sqrt {2{g_A}h} }}{{\sqrt {\dfrac{{2h}}{{{g_A}}}} - \sqrt {\dfrac{{2h}}{{{g_B}}}} }} = \dfrac{{10{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}}}{{2{\rm{ s}}}}\\\ \dfrac{{\sqrt {{g_B}} - \sqrt {{g_A}} }}{{\sqrt {\dfrac{1}{{{g_A}}}} - \sqrt {\dfrac{1}{{{g_B}}}} }} = 5{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}\\\ \sqrt {{g_A}{g_B}} = 5{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}} \end{array}$$ On squaring both sides, we get: $${g_A}{g_B} = 25{\rm{ }}{{{{\rm{m}}^2}} {\left/ {\vphantom {{{{\rm{m}}^2}} {{{\rm{s}}^4}}}} \right. } {{{\rm{s}}^4}}}$$ Therefore, the product of the acceleration due to gravity at the two places A and B is $$25{\rm{ }}{{{{\rm{m}}^2}} {\left/ {\vphantom {{{{\rm{m}}^2}} {{{\rm{s}}^4}}}} \right. } {{{\rm{s}}^4}}}$$. Hence, option (B) is correct. **Note:** We have read the given statements carefully while establishing the mathematical relationships of velocity and time to reach ground. It would be an added advantage to remember the expression of velocity and time when the body is dropped from a certain height under gravity's influence.