Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance).

Step 1: Define the initial parameters.

Let the two balls have masses $m_1$ and $m_2$ (where $m_1 \neq m_2$), both projected vertically upwards with the same initial speed $u$ under negligible air resistance.

Step 2: Determine the acceleration of each ball.

Apply Newton's second law of motion to find the acceleration $a$ under gravity.

$$F = mg$$ $$a = \frac{mg}{m} = g$$

The acceleration is equal to $g$ and is independent of the mass of the ball.

Step 3: Formulate the equations of motion.

Write the kinematic equations for velocity $v(t)$ and vertical displacement $y(t)$ at any time $t$.

$$v(t) = u - gt$$ $$y(t) = ut - \frac{1}{2}gt^2$$

Since mass $m$ does not appear in these equations, both balls have the same velocity and position at every instant.

Step 4: Calculate the maximum height and time to reach the top.

Set the velocity to zero at the maximum height $h$.

$$t_{\text{up}} = \frac{u}{g}$$ $$h = \frac{u^2}{2g}$$

Step 5: Apply conservation of energy to find the return speed.

The kinetic energy at projection equals the potential energy at the maximum height.

$$\frac{1}{2}mu^2 = mgh$$ $$h = \frac{u^2}{2g}$$

Since mass cancels out, both balls return to the point of projection with the same speed.

$$v_{\text{return}} = u$$

Step 6: Determine the total time of flight.

The total time of flight $T$ is independent of mass.

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

Answer: The statement is true; both balls pass through the point of projection in their downward motion with the same speed $u$ and have the same time of flight $T = \dfrac{2u}{g}$, independent of their masses.