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.

To determine when and where two stones, one falling from the top of a tower and the other projected vertically upwards from the ground, will meet, we’ll apply principles of motion under gravity.

Given:
– Height of the tower h₁ = 100 meters
– Initial velocity of the stone projected upwards (u₁) = 25 m/s
– Acceleration due to gravity g = 9.81 m/s² (approximately)

Calculating time for the stone projected upwards to reach maximum height:
Using the equation (v = u + at) where (v) is the final velocity, (u) is the initial velocity, (a) is the acceleration due to gravity, and (t) is time:

At maximum height, the final velocity (v) becomes (0 m/s) momentarily before the stone starts falling back down. So, v = 0, u = 25 m/s, and a = -9.81m/s² (negative due to opposing the upward motion).

v = u + at
0 = 25 – 9.81t
t = 25/9.81
t ≈ 2.55 seconds

Calculating the height reached by the stone projected upwards after 2.55 seconds:

The height (s₂) reached by the stone is calculated using the equation (s = ut + 1/2 at² ):

s₂ = u₁t + 1/2 gt²
s₂ = 25 x 2.55 +1/2 x (-9.81) \times (2.55)²
s₂ ≈ 63.75 meters

Therefore, after approximately 2.55 seconds, the stone projected upwards reaches a height of approximately 63.75 meters.

Conclusion:
Both stones will meet after 2.55 seconds at a height of approximately 63.75 meters above the ground.