Animated Solution for Physics - Kinematics: Two spherical balls having equal masses with radius of $5\text{ cm}$ each are thrown upwards along the same vertical direction at an interval of $3\text{ s}$ with the same initial velocity of $35\text{ m/s}$, then these balls collide at a height of .......... $\text{m}$. (Take, $g=10{\text{ m/s}}^2$)

Imagine standing on an open field, looking straight up into the sky. You throw a ball upwards with a swift initial velocity of $35\text{ m/s}$. You wait exactly $3\text{ seconds}$, and then you throw a second, identical ball upwards with the exact same speed.

It is a classic physics duel. The first ball is already losing its battle against gravity, slowing down as it reaches its peak, while the second ball is just starting its energetic ascent. At some specific height, their paths will cross, and they will collide. Our mission is to find that exact height.

The Master Equation

To track the positions of both balls, we rely on the second equation of motion:

Since gravity is constantly pulling the balls downward, our acceleration $a$ is simply $-g$ (which is $-10{\text{ m/s}}^2$). Let us assume that the collision happens $t$ seconds after the first ball was thrown.

This is where many students make a critical error. The time variable is not the same for both balls.

For the first ball, the time spent in the air is simply $t$. Therefore, its displacement is:

However, the second ball was thrown $3\text{ seconds}$ later. This means it has been flying for $3\text{ seconds}$ less than the first ball. Its time in the air is $(t-3)$. Its displacement equation becomes:

For the two balls to collide, they must be at the exact same height at the exact same time. This means we can equate their displacements:

Now, let's group the similar terms to make the algebra elegant. Bring the linear terms to the left side and the quadratic terms to the right side:

Factor out the common constants:

Notice how beautifully the $t$ cancels out on the left side, leaving us with $35\times 3=105$. On the right side, we can apply the difference of squares formula, $a^2-b^2=(a-b)(a+b)$:

Dividing both sides by $15$:

This tells us that the collision occurs exactly $5\text{ seconds}$ after the first ball was thrown.

Calculating the Height

Now that we have the time of collision, finding the height is a breeze. We simply substitute $t=5\text{ s}$ back into our first displacement equation:

And there we have it! The balls will collide exactly $50\text{ meters}$ above the ground.

Relative Motion

Could we have solved this faster? Absolutely! Let's use the magic of relative motion.

Let's freeze time at $t=3\text{ s}$, the exact moment the second ball is thrown. Where is the first ball? What is its velocity?

At this moment, Ball 2 is at the ground ($S_2=0$) with a velocity of $v_2=35\text{ m/s}$.

Now, let's look at Ball 2 relative to Ball 1. Relative velocity: $v_{rel}=v_2-v_1=35-5=30\text{ m/s}$ (upwards towards Ball 1). Relative acceleration: $a_{rel}=(-g)-(-g)=0$.

Since relative acceleration is zero, Ball 2 approaches Ball 1 at a constant relative speed of $30\text{ m/s}$. The relative distance to cover is $60\text{ m}$.

Time taken to collide = $\frac{\text{Relative Distance}}{\text{Relative Speed}}=\frac{60}{30}=2\text{ s}$.

So, they collide $2\text{ seconds}$ after the second ball is thrown, which means $3+2=5\text{ seconds}$ after the first ball is thrown. The exact same result, achieved in half the time!

Did you notice that the problem mentioned the balls have "equal masses" and a "radius of $5\text{ cm}$"? These were deliberate distractors planted by the examiner!

In basic kinematics (ignoring air resistance), the mass of an object does not affect its acceleration due to gravity. A feather and a bowling ball fall at the same rate in a vacuum. Furthermore, since the collision height ($50\text{ m}$) is vastly larger than the radius of the balls ($0.05\text{ m}$), we can safely treat them as point masses. Always trust your core concepts and don't let extra data intimidate you!