Free falling body

Question

A stone is dropped from the top of the tower 300m high and at the same time another stone is projected vertically upwards with initial velocity 75ms‾¹

Calculate when and where the two stone met.

Answer 1

Typically in kinematics, if you want to solve for one kinematic quantity, you have to know three others, and the process of he problems is to determine what you are trying to find, and what you know.

In this kind of simultaneity question, though, the process is a bit different. The main foundation for solving it is to realize that, for the objects to "meet," they must be in the same position at the same time. So we basically have to set up a kinematic equation for each of the stones, with position and time as the two unknowns.

For the falling stone, we know: v_o (initial velocity), a (acceleration, which is g, downward in free fall), and x_o (initial position, 300 m above the ground). Then we want to throw in displacement (which has final position x in it) and time (t). So we end up with (assuming up is positive, as we often do):

(x - 300 m) = 0 + (1/2)(-g)t² (taking our zero position vertically to be at the ground)

For the rising stone, we know: v_o (+75 m/s), a (g, down, again) and x_o (initial position; at the ground). This gives us an equation of:

x = 75t +(1/2)(-g)t²

(Note: those t's indicate the same time, since both stones are released simultaneously; if they were thrown at different times, we would have to write elapsed time as (t - t_o), with each stone having a different t_o)

Those are two equations with two unknowns: x (final position) and t (time at meeting). Solving for each of those gives you the where and when of the two stones meeting.

If you have any further questions, or would like to check an answer, please let me know. I hope this helps!