Beginner Kinematics Problem

Fix the point where both the particles meet. Let $x_A$ be the distance covered by the particle $A$ and $x_B$ be the distance covered by the particle $B$ .We will use the fact that both particles took the same time $(t)$ to meet.

Consider the motion of $A$ : $x_A=u_At+\dfrac{1}{2}gt^2 \\ x_A=50t + \dfrac{1}{2}(-10)t^2 \\ x_A=50t-5t^2$

Consider the motion of $B$ : $x_B=u_Bt+\dfrac{1}{2}gt^2 \\ x_B=(0)t + \dfrac{1}{2}(10)t^2 \\ x_B=5t^2$

But ,

$x_A+x_B=200 \\ \Rightarrow 50t-5t^2+5t^2=200 \\ \Rightarrow 50t=200 \\ \Rightarrow\boxed{t=4 sec}$

Since both objects are being influenced by gravity, we can look at the movement of the objects relative to the acceleration of gravity itself. Point $A$ is moving up at a constant velocity of $50\text{ m/s}$ while point $B$ is $200\text{ m}$ above point $A$ and stationary. Thus, they take $200\text{ m}\div 50\text{ m/s}=\boxed{4\text{ s}}$ to cross.

My solution is very similar to the one posted by @Nihar Mahajan , even then, I felt like posting, so am I.

Let particle $A$ meet particle $B$ , $x$ m from the ground, hence the Particle $B$ will meet $A$ $x-200$ m from the top.

Let's first take $A$ ,

We know that $x=ut+\dfrac{1}{2}st^{2}$

Substituting the values, we get, $x=50t+\dfrac{1}{2}(-10)t^{2}$

The value of $g$ is -ve because, the particle is moving against gravity.

Or, $x=50-5t^{2}$ $..........(i)$

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Now, let's trace $B$ ,

Again substituting the values, we get, $(200-x)=0t+\dfrac{1}{2}(10)t^{2}$

Here, the value of $g$ is +ve because, the particle is moving towards gravity.

Or, $(200-x)=5t^{2}$

Or, $-x=5t^{2}+200$

Or, $x=-5t^{2}-200$ $..........(ii)$

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Equating $(i)$ and $(ii)$ , $50t-5t^{2}=-5t^{2}-200$

$50t=200$

Therefore, $t=\boxed{4}$

Both particles have the same acceleration ie g downwards so we can use the concept of relative velocity and simply neglect this acceleration. We can fix the position of dropped particle and imagine the thrown particle has a constant velocity 50 m/s upwards. Using time=Distance/speed; t=200/50=4 seconds