Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m s$$^{-1}$$, then these balls collide at a height of _________ m.
(take g = 10 m s$$^{-2}$$)

Let us denote the instant when the first ball is projected upward as $$t = 0$$. Its initial velocity is given as $$u = 35\ \text{m s}^{-1}$$ and it moves against gravity whose magnitude is $$g = 10\ \text{m s}^{-2}$$ (acting downward).

The vertical displacement of any body projected upward with initial speed $$u$$ under constant downward acceleration $$g$$, after a time $$t$$, is given by the well-known kinematic relation

$$s = ut - \tfrac12 g t^{2}.$$

We apply this formula separately to the two balls.

For the first ball, the time elapsed since its launch is simply $$t$$ seconds. Hence its height above the point of projection is

$$\begin{aligned} y_{1} &= u t - \tfrac12 g t^{2} \\ &= 35\,t - 5\,t^{2}. \quad -(1) \end{aligned}$$

The second ball is thrown upward exactly $$3\ \text{s}$$ after the first one. Therefore, when the clock shows $$t$$ seconds after the first throw, the second ball has been in flight for only $$(t-3)$$ seconds. (Of course, this expression is meaningful only when $$t \ge 3\ \text{s}$$, i.e. after the second ball has actually been launched.) Using the same formula, its height is

$$\begin{aligned} y_{2} &= u\,(t-3) - \tfrac12 g\,(t-3)^{2} \\ &= 35\,(t-3) - 5\,(t-3)^{2}. \quad -(2) \end{aligned}$$

The collision occurs when both balls are at the same height, so we set $$y_{1} = y_{2}$$ using equations (1) and (2).

$$\begin{aligned} 35\,t - 5\,t^{2} &= 35\,(t-3) - 5\,(t-3)^{2}. \end{aligned}$$

First we expand the right-hand side:

$$\begin{aligned} 35\,(t-3) &= 35t - 105,\\ (t-3)^{2} &= t^{2} - 6t + 9,\\ -5\,(t-3)^{2} &= -5t^{2} + 30t - 45. \end{aligned}$$

Adding these components, the right-hand side becomes

$$\bigl(35t -105\bigr) + \bigl(-5t^{2} + 30t - 45\bigr) = 65t - 5t^{2} - 150.$$

Equating both sides we have

$$35t - 5t^{2} = 65t - 5t^{2} - 150.$$

The terms $$-5t^{2}$$ appear on both sides and cancel out immediately, leaving

$$35t = 65t - 150.$$

Transposing the terms gives

$$35t - 65t = -150 \quad\Longrightarrow\quad -30t = -150.$$

Dividing by $$-30$$ we find

$$t = 5\ \text{s}.$$

Thus, the collision happens $$5\ \text{s}$$ after the first ball was projected, which also means $$5 - 3 = 2\ \text{s}$$ after the second ball was projected.

To obtain the height at which they meet, we substitute $$t = 5\ \text{s}$$ into equation (1) (the expression for the first ball, though either expression would give the same result):

$$\begin{aligned} y &= 35\,(5) - 5\,(5)^{2} \\ &= 175 - 5 \times 25 \\ &= 175 - 125 \\ &= 50\ \text{m}. \end{aligned}$$

Therefore, the two spherical balls collide at a height of $$50\ \text{m}$$ above the point of projection.

So, the answer is $$50\ \text{m}.$$