A stone is allowed to fall from the top of a tower 100 m high ... - Science

Given:

• Height of the tower (H) = 100 \text{ m}

• Initial velocity of falling stone (u_1) = 0 \text{ m/s}

• Initial velocity of rising stone (u_2) = 25 \text{ m/s}

• Acceleration due to gravity (g) = 9.8 \text{ m/s}^2

1. When will they meet (Time):
Let the two stones meet after time t.
The distance covered by the falling stone + The distance covered by the rising stone = Total Height.
(u_1t + \frac{1}{2}gt^2) + (u_2t - \frac{1}{2}gt^2) = 100
(0 + \frac{1}{2}gt^2) + (25t - \frac{1}{2}gt^2) = 100
Canceling \frac{1}{2}gt^2 from both sides:
25t = 100
t = \frac{100}{25}
t = 4 \text{ s}

2. Where will they meet (Height):
We calculate the height covered by the rising stone in 4 seconds.

h = u_2t - \frac{1}{2}gt^2
h = 25(4) - \frac{1}{2} \times 9.8 \times (4)^2
h = 100 - \frac{1}{2} \times 9.8 \times 16
h = 100 - (9.8 \times 8)
h = 100 - 78.4
h = 21.6 \text{ m}

Answer: The two stones will meet 4 seconds after the start at a height of 21.6 m from the ground.