A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.

Bobbie Comstock 2021-12-17 Answered

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch ayt,vyt, and y-t graphs for the motion.

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Expert Answer

Fasaniu
Answered 2021-12-18 Author has 46 answers

Step 1
1Concepts and Principles
1- Particle under constant acceleration: If a particle moves in a straight line with a constant acceleration ax, its motion is described by kinematics equations, from which we will use the following equation:
(1)vxf=vxi+ayt
(2)xf=xi+vξt+12axt2
2- The sulution to a quadration in the from: ax2+bx+x=0 is given by the quadrat formula:
x=b±b24ac2a
Step 2
2 Given Data
Let the +y-direction be upwards and the initial position of the sand bag be at yi=0.
vyi(initial velocity of the bag)=5ms
yf(position of the bag when it reaches the ground)=40m
ay(acceleration due to gravity)=9.8ms2
Step 3
3 Required Data
In part (a), we are asked to find the position and the velocity of the sandbag at t=0.25 s and t=1 s.
In part(b), we are asked to find the time taken by the bag to reach the ground.
In part(c), we are asked to find the speed with which the bag strikes the ground.
In part(d), we are asked to find the maximum height the bag reaches above the ground.
In part(e), we are asked to sketch ayt,vyt,and y task graphs for the motion of the sand bag.
Step 4
4 Solution
(a)
Model the sand bag as a particle under constant acceleration and apply Equations (1) and (2) to find the final velocity and final position of the sand bag as a function of time:
vyf=vyi+ayt
yf=yi+vyit+12ayt2
For t=0.25 s:
vyf=5ms+(9.8ms2)(0.25s)=2.55ms
yf=0+(5ms)(0.25s)+12(9.8ms2)(0.25s)2=0.944m
Therefore, at t=0.25 s, the sand bag is moving with an upward velocity of magnitude 2.55 m/s and is 40 m+0.944 m=40.9 m above the ground.
For t=0.1 s:
vyf=5ms+(9.8ms2)(1s)=4.8ms
yf=0+(5ms)(1s)+12(9.8ms2)(1s)2=0.1m
Therefore, at t=1 s, the sand bag is moving with an downward velocity of magnitude 4.8 m/s and is 40 m+0.1 m=40.1 m above the ground.

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Rita Miller
Answered 2021-12-19 Author has 28 answers

Step 5
(b)
The sand bag strikes the ground when its final position is yf=40m. From Equation (2):
yf=yi+vyit+12ayt2
40m=0+(5ms)t+12(9.8ms2)t2
40m=(5ms)t(4.9ms2)t2
(4.9ms2)t2(5ms)t40m=0
Solve this equation for t using the quadratic formula from Equation (3), for a=4.9ms2,b=5ms, and c=40m:
t=b±b24ac2a
t=(5ms)±(5ms)24(4.9ms2)(40m)2(4.9ms2)
=3.41 s or -2.39 s
A negative value for time is phisically unacceptable. Therefore, the time take for the sand bag to reach the ground is t=3.41 s.
Step 6
(c)
The velocity of the bag before it atrikes the ground is fround from Equation (l):
vyf=vyi+ayt
Where t is the time taken for the sand bag to reach the ground which we found in part (b). Subsitute numerical values:
vyf=5ms+(9.8ms2)(3.41s)=28.4ms
The negative values indicates that the direction of the velocity ia downwads.
Step 7
(d)
The bag reaches maximum heightwhen its final velosity vyf=0. From Equation (2), we find the time taken by the bag to reach this maximum height:
vyf=vyi+ayt
0=5 m/s+(9.8 m/s2)t
t=5ms9.8ms2=0.51s
They by using Equation (2), we can find the maximum height:
y=y0+vyit+12ayt2
y=05ms(0.51s)+12(9.8ms2)(0.51s)2
y=1.28 m
Therefore, the maximum height of the sand bag is 40 m+1.28 m=41.3 m.

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Don Sumner
Answered 2021-12-27 Author has 35 answers

Step 8
(e)
The ayt graph is a horithontal line since the acceleration is constant at ay=9.8m/s2
image
Step 9
The velosity oa the sand bag is a function of time from Eguation (l) as follows:
vy=vyi+ayt=5m/s(9.8m/s2)t
Therefore, the vyt graphs looks like:
image

Step 10
The y - t graph look like (let y=0 at the ground level):
image
Result
(a)At t=0.25 s, the sand bag is moving with an upward velocity of magnitude 2.55 m/s and is 40.9 m above the ground. At t=1 s, the sand bag is moving with an downwoard velosity of magnitude 4.8 m/s and is 40.1 m above the ground.
(b)t=3.41 s
(c)vyt=28.4 m/s
(d)The maximum height of the sand is 41.3 m.
(e)Click for graphs

 

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