Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle θ, both of them stay in our view for the same period. The height attained by the two balls are in the ratio

The correct option is 1 : 1.

Let us break down the problem step-by-step to understand the relation between the time a projectile stays in the air (time of flight) and the maximum height it reaches.

Step 1: Understand the motion and the parameters
Pankaj throws his ball of mass m vertically upwards. Let its initial vertical velocity be $u_{y1}$.
Sudhir throws his ball of mass 2m at an angle $\theta$ to the horizontal. Let its initial vertical velocity component be $u_{y2}$.
Note that the mass of the balls does not affect their gravitational acceleration or motion under gravity, as gravity acts equally on all masses near the Earth's surface (ignoring air resistance).

Step 2: Relate the time of flight to the initial vertical velocity
The time of flight (T) for any object projected near the Earth's surface depends solely on its initial vertical velocity component ($u_y$) and the acceleration due to gravity (g). It is given by the formula:

$T=\frac{2u_y}{g}$

We are given that both balls stay in our view for the same period of time, which means their times of flight are equal ($T_1=T_2$):

$\frac{2u_{y1}}{g}=\frac{2u_{y2}}{g}$

From this relation, we find that their initial vertical components of velocity must be equal:

$u_{y1}=u_{y2}$

Step 3: Relate the maximum height to the initial vertical velocity
The maximum vertical height (H) attained by a projected body also depends only on the initial vertical component of its velocity ($u_y$) and is expressed as:

$H=\frac{u_y^2}{2g}$

Let $H_1$ be the height attained by Pankaj's ball and $H_2$ be the height attained by Sudhir's ball. The ratio of their heights is:

$\frac{H_1}{H_2}=\frac{\left(\frac{u_{y1}^2}{2g}\right)}{\left(\frac{u_{y2}^2}{2g}\right)}=\frac{u_{y1}^2}{u_{y2}^2}$

Since we established in Step 2 that $u_{y1}=u_{y2}$, we substitute this into the ratio:

$\frac{H_1}{H_2}=\frac{u_{y1}^2}{u_{y1}^2}=1$

Thus, the ratio of the heights attained by the two balls is 1 : 1.