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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 ms-2. He reaches the ground with a speed of 3 ms-1. At what height, did he bail out ? [AIEEE 2005]
- a)91 m
- b)182 m
- c)293 m
- d)111 m
Correct answer is option 'C'. Can you explain this answer?
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A parachutist after bailing out falls 50 m without friction. When para...
Parachute bails out at height H from ground. Velocity at A:
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A parachutist after bailing out falls 50 m without friction. When para...
To solve this problem, we can use the equations of motion for the parachutist's free fall and the deceleration caused by the parachute.
Let's break down the problem step by step:
1. Initial velocity:
The parachutist starts free falling with an initial velocity of zero.
2. Free fall distance:
The parachutist falls a distance of 50 m without any friction or deceleration.
3. Parachute opening:
When the parachute opens, the parachutist experiences a deceleration of 2 m/s².
4. Final velocity:
The parachutist reaches the ground with a final velocity of 3 m/s.
Now, let's calculate the height at which the parachutist bailed out:
1. Calculate the time taken during free fall:
We can use the equation of motion: s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.
For free fall, s = 50 m, u = 0 m/s, and a = 9.8 m/s² (acceleration due to gravity). Plugging in these values, we get:
50 = 0*t + (1/2)*9.8*t²
Simplifying the equation, we get:
4.9t² = 50
t² = 50/4.9
t ≈ 3.19 s (taking the positive root)
2. Calculate the time taken during parachute deceleration:
We can again use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
For parachute deceleration, v = 3 m/s, u = 0 m/s, and a = -2 m/s² (negative because it is deceleration). Plugging in these values, we get:
3 = 0 + (-2)*t
t = 3/2
t = 1.5 s
3. Calculate the height at which the parachute opened:
During the deceleration, the parachutist covered a distance equal to the distance fallen during free fall. Therefore, the height at which the parachute opened is:
h = s - ut - (1/2)at²
h = 50 - 0*1.5 - (1/2)*(-2)*(1.5)²
h = 50 - 0 - (-1.5)
h = 51.5 m
Therefore, the parachutist bailed out at a height of approximately 51.5 m. However, none of the given options match this answer. It seems that there might be an error in the provided answer choices.
Let's break down the problem step by step:
1. Initial velocity:
The parachutist starts free falling with an initial velocity of zero.
2. Free fall distance:
The parachutist falls a distance of 50 m without any friction or deceleration.
3. Parachute opening:
When the parachute opens, the parachutist experiences a deceleration of 2 m/s².
4. Final velocity:
The parachutist reaches the ground with a final velocity of 3 m/s.
Now, let's calculate the height at which the parachutist bailed out:
1. Calculate the time taken during free fall:
We can use the equation of motion: s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.
For free fall, s = 50 m, u = 0 m/s, and a = 9.8 m/s² (acceleration due to gravity). Plugging in these values, we get:
50 = 0*t + (1/2)*9.8*t²
Simplifying the equation, we get:
4.9t² = 50
t² = 50/4.9
t ≈ 3.19 s (taking the positive root)
2. Calculate the time taken during parachute deceleration:
We can again use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
For parachute deceleration, v = 3 m/s, u = 0 m/s, and a = -2 m/s² (negative because it is deceleration). Plugging in these values, we get:
3 = 0 + (-2)*t
t = 3/2
t = 1.5 s
3. Calculate the height at which the parachute opened:
During the deceleration, the parachutist covered a distance equal to the distance fallen during free fall. Therefore, the height at which the parachute opened is:
h = s - ut - (1/2)at²
h = 50 - 0*1.5 - (1/2)*(-2)*(1.5)²
h = 50 - 0 - (-1.5)
h = 51.5 m
Therefore, the parachutist bailed out at a height of approximately 51.5 m. However, none of the given options match this answer. It seems that there might be an error in the provided answer choices.
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A parachutist after bailing out falls 50 m without friction. When para...
Answer C is correct
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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 ms-2. He reaches the ground with a speed of 3 ms-1.At what height, did he bail out ? [AIEEE 2005]a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer?
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