Answer to Question #166176 in Physics for gibson

Question

A first student drops a ball point pen from the 4th floor of the hostel 400m high and at the same time another is projected vertically upward from the ground with a velocity of 100m/s. Find when and where two pens will meet.

Answer 1

Let's choose the upwards as the positive direction. Let's write the vertical displacement for the first and second ball:

$$y_1=y_0+v_{01}t-\dfrac{1}{2}gt^2,$$ $$y_2=v_{02}t-\dfrac{1}{2}gt^2.$$

The two balls will meeting when $y_1=y_2$:

$$y_0+0-\dfrac{1}{2}gt^2=v_{02}t-\dfrac{1}{2}gt^2,$$ $$y_0=v_{02}t,$$ $$t=\dfrac{y_0}{v_{02}}=\dfrac{400\ m}{100\ \dfrac{m}{s}}=4\ s.$$

As we find the time at what the two balls were met, we can find the height:

$$y=400\ m-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(4\ s)^2=321.6\ m.$$