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.