Question

Two objects X and Y are thrown upwards simultaneously with the same speed. The mass of X is greater than that of Y. the air exerts equal resistive force on the two objects, then,
(A) X will be higher than Y
(B) Y will be higher than X
(C) The two objects will reach the same height
(D) Cannot say

Answer 1

Hint : The objects X and Y have the same initial velocity and will also have the same velocity at maximum height which is zero. We need to investigate the acceleration of the two masses.

Formula used: In this solution we will be using the following formulae;
${F_{net}} = ma$ where $m$ is the mass of an object ${F_{net}}$ is the net force acting on the object, $a$ is the acceleration of the object.
${v^2} = {u^2} + 2as$ where $v$ is the final velocity of a body, $u$ is the initial velocity $a$ is the acceleration of the body and is the distance covered within that period.

Complete step by step answer:
To investigate which of these two masses will have a higher maximum height, we must know the difference between their accelerations.
Performing a Newton’s second law analysis on the two masses we have
${F_{net}} = ma$ where $m$ is the mass of an object ${F_{net}}$ is the net force acting on the object, $a$ is the acceleration of the object.
$\Rightarrow {F_{x,net}} = {m_x}g + f = {m_x}{a_x}$ where ${F_{x,net}}$ is the net force on mass X, ${m_x}$ is the mass $g$ is the acceleration due to gravity and $f$ is the resistive force on the mass.
Similarly,
${F_{y,net}} = {m_y}g + f = {m_y}{a_y}$
Hence, their accelerations are
${a_x} = g + \dfrac{f}{{{m_x}}}$ And
${a_y} = g + \dfrac{f}{{{m_y}}}$
Now since ${m_x} > {m_y}$ then ${a_y} > {a_x}$
Now, from the equation of motion
${v^2} = {u^2} + 2as$ where $v$ is the final velocity of a body, $u$ is the initial velocity $a$ is the acceleration of the body and is the distance covered within that period
We know at maximum height, they both have zero velocity. Accordingly, the initial velocities are equal, hence
$0 = {u^2} - 2{a_x}{s_x}$ (assuming upward is positive), and
$0 = {u^2} - 2{a_y}{s_y}$
Hence,
${s_x} = \dfrac{{{u^2}}}{{2{a_x}}}$ and
${s_y} = \dfrac{{{u^2}}}{{2{a_y}}}$
Since, ${a_y} > {a_x}$ then ${s_x} > {s_y}$
Hence, X will be higher than Y.
The correct option is A.

Note:
For clarity, though it may be tempting to subtract the resistive force from the weight of the body as in ${F_{y, net}} = {m_y}g - f = {m_x}a$ , it would be wrong. This is because both the weight of the body and the resistive force are acting downward while the object is going up.