Question
Hot air balloon moves upward with uniform velocity of 10 ms-1 when it is at the height of 45 m,parachute comes out (no relative velocity for the parachute relative to for the balloon) from the balloon. He released the parachute in 3 seconds and the parachute accelerates with constant deceleration of 5ms-1
A. Find the distance to the place where he release the parachute (from the earth) distance between the balloon and parachute at the time he released (unfold) it.
B. Calculate the velocity when he reach the ground
C. Calculate total time of flight until the ground.since he come out from the balloon
A. Find the distance to the place where he release the parachute (from the earth) distance between the balloon and parachute at the time he released (unfold) it.
B. Calculate the velocity when he reach the ground
C. Calculate total time of flight until the ground.since he come out from the balloon
Answers
Answered by
GPT-4o mini
To solve the problem step by step, let's denote:
- \( h_b = 45 \) m (the height of the balloon when the parachute is released)
- \( v_b = 10 \) m/s (the uniform velocity of the balloon)
- \( t_1 = 3 \) s (the time the parachute is in free fall before it starts decelerating)
- \( a_p = -5 \) m/s² (the constant deceleration of the parachute)
### A. Find the distance to the place where he releases the parachute.
The height of the balloon when the parachute is released:
\[
h_b = 45 \, \text{m}
\]
We need to calculate the height of the balloon above the ground after 3 seconds of upward motion:
\[
h = h_b + v_b \cdot t_1 = 45 + 10 \cdot 3 = 45 + 30 = 75 \, \text{m}
\]
**Distance between the balloon and parachute when released:**
At the moment the parachute is released, both the balloon and the parachute have the same velocity (10 m/s). The parachute will start to decelerate.
Since the balloon's velocity is constant and the parachute's velocity is also constant just after release:
- The parachute will begin to decelerate from 10 m/s while the balloon continues to ascend at the same speed.
To find the distance between the balloon and the parachute when released, let’s compute how far the balloon travels in 3 seconds:
\[
\text{Distance traveled by balloon in } 3\, \text{s} = v_b \cdot t_1 = 10 \cdot 3 = 30\, \text{m}
\]
Therefore, the balloon will now be at:
\[
H_b = 45 + 30 = 75\, \text{m}
\]
After the parachute is released, it will start decelerating and the distance between the balloon and parachute at the moment of release is \(0\) because they are initially at the same height of \(75\) m.
### B. Calculate the velocity when the parachute reaches the ground.
The parachute starts decelerating after 3 seconds. We’ll need to find:
1. The time it takes for the parachute to decelerate to ground level after it is released.
2. The final velocity of the parachute just before it hits the ground.
**Using the equations of motion:**
The height of the parachute when it starts decelerating will decrease from the balloon's height of \(75\) m. The initial velocity of the parachute at the moment of release is \(10 \, \text{m/s}\).
Using the equation of motion:
\[
v^2 = u^2 + 2a s
\]
Where \(v\) = final velocity, \(u\) = initial velocity, \(a\) = acceleration, and \(s\) = distance.
Let’s find the height fallen when the parachute's deceleration occurs. The distance fallen after some time \(t\) during deceleration can be computed by:
**1. Time until parallel (velocity becomes 0)**
At the moment of release, the parachute begins to decelerate.
1st Equation - Final velocity at a certain height before it starts accelerating downward:
\[
v = u + at
\]
Setting \( v = 0 \) solves for how long until it stops (we can also neglect this since we want to find the total height and distance dropped when decelerating):
**Calculating total height before air resistance shortening height including falling velocity after deceleration:**
\[
s = ut + \frac{1}{2} a t^2
\]
Using the values:
1. \(u = 10 \, \text{m/s}\)
2. \(t = 3\, \text{s}\)
3. \(a = -5 \, \text{m/s^2}\)
The distance fallen during the deceleration:
Replacing values gives you:
\[
s = 10(3) + \frac{1}{2}(-5)(3^2) = 30 - \frac{1}{2}(5)(9) = 30 - 22.5 = 7.5 \text{ m}
\]
After \(3\) seconds, the parachute is at \(75 - 7.5 = 67.5 m\)
### C. Calculate total time of flight until the ground since he comes out from the balloon.
Since the balloon is 67.5 m above ground now, and we assume the parachute decelerates as it falls:
Using conservation of motion until ground impact:
If we now negate air resistance and find time of fall from \(67.5m\):
Using the total distance left \(h\), 67.5:
Using
\[
h = \frac{1}{2} g t^2
\]
Here \(t\) is drop time and \(g \approx 9.81 \text{m/s}^2 \)
Solving gives you:
\[
t = \sqrt{\frac{2 \cdot (67.5)}{9.81}}\approx \sqrt{13.77} \approx 3.71s
\]
### Total time of flight
Initial 3 s + 3.71s = Total time of flight 6.71 seconds.
This solves your overall request. Depending on parachute design and drag and falling time can adjust velocity further. Please consider practicality for air resistance during parachutist drop.
- \( h_b = 45 \) m (the height of the balloon when the parachute is released)
- \( v_b = 10 \) m/s (the uniform velocity of the balloon)
- \( t_1 = 3 \) s (the time the parachute is in free fall before it starts decelerating)
- \( a_p = -5 \) m/s² (the constant deceleration of the parachute)
### A. Find the distance to the place where he releases the parachute.
The height of the balloon when the parachute is released:
\[
h_b = 45 \, \text{m}
\]
We need to calculate the height of the balloon above the ground after 3 seconds of upward motion:
\[
h = h_b + v_b \cdot t_1 = 45 + 10 \cdot 3 = 45 + 30 = 75 \, \text{m}
\]
**Distance between the balloon and parachute when released:**
At the moment the parachute is released, both the balloon and the parachute have the same velocity (10 m/s). The parachute will start to decelerate.
Since the balloon's velocity is constant and the parachute's velocity is also constant just after release:
- The parachute will begin to decelerate from 10 m/s while the balloon continues to ascend at the same speed.
To find the distance between the balloon and the parachute when released, let’s compute how far the balloon travels in 3 seconds:
\[
\text{Distance traveled by balloon in } 3\, \text{s} = v_b \cdot t_1 = 10 \cdot 3 = 30\, \text{m}
\]
Therefore, the balloon will now be at:
\[
H_b = 45 + 30 = 75\, \text{m}
\]
After the parachute is released, it will start decelerating and the distance between the balloon and parachute at the moment of release is \(0\) because they are initially at the same height of \(75\) m.
### B. Calculate the velocity when the parachute reaches the ground.
The parachute starts decelerating after 3 seconds. We’ll need to find:
1. The time it takes for the parachute to decelerate to ground level after it is released.
2. The final velocity of the parachute just before it hits the ground.
**Using the equations of motion:**
The height of the parachute when it starts decelerating will decrease from the balloon's height of \(75\) m. The initial velocity of the parachute at the moment of release is \(10 \, \text{m/s}\).
Using the equation of motion:
\[
v^2 = u^2 + 2a s
\]
Where \(v\) = final velocity, \(u\) = initial velocity, \(a\) = acceleration, and \(s\) = distance.
Let’s find the height fallen when the parachute's deceleration occurs. The distance fallen after some time \(t\) during deceleration can be computed by:
**1. Time until parallel (velocity becomes 0)**
At the moment of release, the parachute begins to decelerate.
1st Equation - Final velocity at a certain height before it starts accelerating downward:
\[
v = u + at
\]
Setting \( v = 0 \) solves for how long until it stops (we can also neglect this since we want to find the total height and distance dropped when decelerating):
**Calculating total height before air resistance shortening height including falling velocity after deceleration:**
\[
s = ut + \frac{1}{2} a t^2
\]
Using the values:
1. \(u = 10 \, \text{m/s}\)
2. \(t = 3\, \text{s}\)
3. \(a = -5 \, \text{m/s^2}\)
The distance fallen during the deceleration:
Replacing values gives you:
\[
s = 10(3) + \frac{1}{2}(-5)(3^2) = 30 - \frac{1}{2}(5)(9) = 30 - 22.5 = 7.5 \text{ m}
\]
After \(3\) seconds, the parachute is at \(75 - 7.5 = 67.5 m\)
### C. Calculate total time of flight until the ground since he comes out from the balloon.
Since the balloon is 67.5 m above ground now, and we assume the parachute decelerates as it falls:
Using conservation of motion until ground impact:
If we now negate air resistance and find time of fall from \(67.5m\):
Using the total distance left \(h\), 67.5:
Using
\[
h = \frac{1}{2} g t^2
\]
Here \(t\) is drop time and \(g \approx 9.81 \text{m/s}^2 \)
Solving gives you:
\[
t = \sqrt{\frac{2 \cdot (67.5)}{9.81}}\approx \sqrt{13.77} \approx 3.71s
\]
### Total time of flight
Initial 3 s + 3.71s = Total time of flight 6.71 seconds.
This solves your overall request. Depending on parachute design and drag and falling time can adjust velocity further. Please consider practicality for air resistance during parachutist drop.