Question

A balloon is ascending vertically with a acceleration of 0.2m/s square .two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5 sec. After the second stone is released

Answer 1

Answer on Question #42670-Physics-Mechanics-Kinematics-Dynamics

A balloon is ascending vertically with an acceleration of $0.2\mathrm{m/s}$ square .two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5 sec. After the second stone is released

Solution

$$t _ {1} = 2 s, t _ {2} = 1. 5 s, a = 0. 2 \frac {m}{s ^ {2}}.$$

Let $V$ be the velocity of the balloon when the first stone is dropped from A, the velocity of the balloon, when the second stone is dropped from B, is

$$V _ {1} = V + a t _ {1} = V + 0. 2 \cdot 2 = V + 0. 4 \frac {m}{s}.$$

Then

$$A B = V t _ {1} + \frac {a t _ {1} ^ {2}}{2} = 2 V + 0. 2 \cdot \frac {2 ^ {2}}{2} = 2 V + 0. 4 m.$$

Both these particles will start moving upwards from A and B with these velocities $V$ and $V_{1}$ respectively.

After 3.5 seconds when the first stone was dropped, i.e. 1.5 seconds when the second stone was dropped, let the two stones be at C and D respectively. Obviously D is above C and

$$A C = 3. 5 V - \frac {1}{2} g \cdot 3. 5 ^ {2}.$$ $$B D = 1. 5 V _ {1} - \frac {1}{2} g \cdot 1. 5 ^ {2}.$$

Distance between the two stones at this time

$$\begin{array}{l} C D = A D - A C = (A B + B D) - A C = \left(2 V + 0. 4 + 1. 5 (V + 0. 4) - \frac {1}{2} g \cdot 1. 5 ^ {2}\right) - \left(3. 5 V - \frac {1}{2} g \cdot 3. 5 ^ {2}\right) \\ = 1 + 5 g = 5 0 m. \\ \end{array}$$

Answer: $50\mathrm{m}$

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