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An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is:
- a)1200 m
- b)0.33 km
- c)3.33 km
- d)33 km
Correct answer is option 'C'. Can you explain this answer?
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An aeroplane is flying horizontally with a velocity of 600 km/h at a h...
Horizontal displacement of the bomb AB = Horizontal velocity × time available
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An aeroplane is flying horizontally with a velocity of 600 km/h at a h...
Given data:
Velocity of the airplane, v = 600 km/h
Height of the airplane, h = 1960 m
To find:
The distance AB, i.e., the horizontal distance covered by the bomb before it strikes the ground.
Assumptions:
1. The horizontal velocity of the bomb is the same as the airplane's velocity.
2. There is no air resistance acting on the bomb.
Analysis:
1. The time taken by the bomb to reach the ground can be calculated using the equation of motion: s = ut + 0.5at², where s is the vertical distance, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time taken.
- The vertical distance covered by the bomb, s = h = 1960 m
- The initial vertical velocity, u = 0 (as the bomb is released from rest)
- The acceleration due to gravity, a = 9.8 m/s² (assuming no air resistance)
- Substituting the values, we get: 1960 = 0 + 0.5 × 9.8 × t²
- Solving the equation, we find t = √(2h/a) = √(2 × 1960/9.8) = √(400) = 20 seconds
2. The horizontal distance covered by the bomb can be calculated using the formula: distance = velocity × time.
- The horizontal velocity of the bomb is the same as the airplane's velocity, v = 600 km/h.
- Converting the velocity to m/s: v = 600 × (1000/3600) = 600 × (5/18) = 500/3 m/s
- Substituting the values, we get: distance = (500/3) × 20 = 10000/3 = 3333.33 meters = 3.33 km
Therefore, the distance AB is 3.33 km (option C).
Velocity of the airplane, v = 600 km/h
Height of the airplane, h = 1960 m
To find:
The distance AB, i.e., the horizontal distance covered by the bomb before it strikes the ground.
Assumptions:
1. The horizontal velocity of the bomb is the same as the airplane's velocity.
2. There is no air resistance acting on the bomb.
Analysis:
1. The time taken by the bomb to reach the ground can be calculated using the equation of motion: s = ut + 0.5at², where s is the vertical distance, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time taken.
- The vertical distance covered by the bomb, s = h = 1960 m
- The initial vertical velocity, u = 0 (as the bomb is released from rest)
- The acceleration due to gravity, a = 9.8 m/s² (assuming no air resistance)
- Substituting the values, we get: 1960 = 0 + 0.5 × 9.8 × t²
- Solving the equation, we find t = √(2h/a) = √(2 × 1960/9.8) = √(400) = 20 seconds
2. The horizontal distance covered by the bomb can be calculated using the formula: distance = velocity × time.
- The horizontal velocity of the bomb is the same as the airplane's velocity, v = 600 km/h.
- Converting the velocity to m/s: v = 600 × (1000/3600) = 600 × (5/18) = 500/3 m/s
- Substituting the values, we get: distance = (500/3) × 20 = 10000/3 = 3333.33 meters = 3.33 km
Therefore, the distance AB is 3.33 km (option C).
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An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is:a)1200 mb)0.33 kmc)3.33 kmd)33 kmCorrect answer is option 'C'. Can you explain this answer?
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