An aeroplane is flyind horizontally at a height of 2 km and with a velocity of 720 km/h. A bag containing ration is to be dropped to the Jawans on the ground then find out
(i) How far from the Jawans should the bag be released so that it falls directly over them
(ii) Time taken by bag to reach the ground
(iii) Velocity of bag when it reach the ground.
(iv) At which angle the bag will reaches the ground.
(v) What path of bag will be appeared to pilot
(vi) What path of bag will be appeared to Jawans.
(vii) Position of aeroplane when bag reaches to the Jawans.`(g=10m//s^(2))`

Answer

To solve the problem step by step, we will break down each part of the question systematically.

### Given Data:
- Height of the aeroplane (h) = 2 km = 2000 m
- Velocity of the aeroplane (v) = 720 km/h = 200 m/s (after conversion)
- Acceleration due to gravity (g) = 10 m/s²

### (i) Distance from the Jawans to release the bag
To find out how far from the Jawans the bag should be released, we first need to calculate the time taken by the bag to reach the ground.

#### Step 1: Calculate the time taken to reach the ground (t)
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here,
- \( s = 2000 \, \text{m} \) (downward, so we take it as positive),
- \( u = 0 \) (initial vertical velocity),
- \( a = g = 10 \, \text{m/s}^2 \).

Substituting the values:
\[ 2000 = 0 + \frac{1}{2} \times 10 \times t^2 \]
\[ 2000 = 5t^2 \]
\[ t^2 = \frac{2000}{5} = 400 \]
\[ t = \sqrt{400} = 20 \, \text{s} \]

#### Step 2: Calculate the horizontal distance (d)
Now, we can find the horizontal distance using the formula:
\[ d = v \times t \]
Where \( v = 200 \, \text{m/s} \) and \( t = 20 \, \text{s} \):
\[ d = 200 \times 20 = 4000 \, \text{m} \]

### (ii) Time taken by the bag to reach the ground
The time taken by the bag to reach the ground is already calculated as:
\[ t = 20 \, \text{s} \]

### (iii) Velocity of the bag when it reaches the ground
To find the final velocity of the bag just before it hits the ground, we can use the equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( u = 0 \) (initial vertical velocity),
- \( a = g = 10 \, \text{m/s}^2 \),
- \( s = 2000 \, \text{m} \).

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 2000 \]
\[ v^2 = 40000 \]
\[ v = \sqrt{40000} = 200 \, \text{m/s} \]

### (iv) Angle at which the bag will reach the ground
The angle \( \theta \) can be calculated using:
\[ \tan \theta = \frac{v_y}{v_x} \]
Where:
- \( v_y = 200 \, \text{m/s} \) (vertical component),
- \( v_x = 200 \, \text{m/s} \) (horizontal component).

Thus,
\[ \tan \theta = \frac{200}{200} = 1 \]
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### (v) Path of the bag as seen by the pilot
To the pilot, the bag appears to move in a straight line downwards because it retains the horizontal velocity of the plane at the moment of release.

### (vi) Path of the bag as seen by the Jawans
To the Jawans on the ground, the bag follows a parabolic path due to the influence of gravity acting on it while it also moves horizontally.

### (vii) Position of the aeroplane when the bag reaches the Jawans
When the bag reaches the Jawans, the aeroplane will be directly above the point where the bag was dropped, at a height of 2 km.

### Summary of Answers:
1. Distance from the Jawans to release the bag: **4000 m**
2. Time taken by the bag to reach the ground: **20 s**
3. Velocity of the bag when it reaches the ground: **200 m/s**
4. Angle at which the bag will reach the ground: **45°**
5. Path of the bag as seen by the pilot: **Straight line**
6. Path of the bag as seen by the Jawans: **Parabolic**
7. Position of the aeroplane when the bag reaches the Jawans: **Directly above the Jawans at a height of 2 km**