A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Sure, let's solve this step by step.

1. First, let's consider the stone falling from the tower. The distance it falls (d1) can be calculated using the equation of motion: d1 = ut + 0.5gt^2, where u is the initial velocity, g is the acceleration due to gravity, and t is the time. Since the stone is simply falling, its initial velocity is 0. So, the equation becomes: d1 = 0.5gt^2.

2. Now, let's consider the stone being thrown upwards. The distance it travels (d2) can also be calculated using the equation of motion: d2 = ut - 0.5gt^2, where u is the initial velocity (25 m/s), g is the acceleration due to gravity, and t is the time.

3. The two stones will meet when d1 + d2 equals the height of the tower. So, we can set up the equation: 0.5gt^2 + (25t - 0.5gt^2) = 100.

4. Simplifying this equation gives: 25t = 100, so t = 100 / 25 = 4 seconds. So, the two stones will meet 4 seconds after they are released.

5. To find out where the stones will meet, we can substitute t = 4 seconds into either of the two distance equations. If we use the first one (for the stone falling from the tower), we get: d1 = 0.5g(4^2) = 0.5 * 9.8 * 16 = 78.4 meters. So, the stones will meet 78.4 meters above the ground.