(a) The maximum altitude of the rocket.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

81 SSM Figure 29-84 shows a cross section of an infinite conducting sheet carrying a current per unit x-length of 2; the current emerges perpendicularly out of the page. (a) Use the Biot-Savart law and symmetry to show that for all points B P P. BD P' Figure 29-84 Problem 81. x P above the sheet and all points P' below it, the magnetic field B is parallel to the sheet and directed as shown. (b) Use Ampere's law to prove that B = ½µλ at all points P and P'.

What All equations of Ountum physics?

Please rewrite the rules of Quantum mechanics?

Answer 2

Question

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Answer 6

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

Answer 7

Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Answer 8

Expert Solution

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7 m$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

a₁ = $2.00 m/s2$

a₂ = $3.00 m/s2$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

$H=ut+12at2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

$H1=ut1+12a1t12=0×4+12×2×162=256 m$

At the end of first stage speed is,

$v1=u+a1t1=0+2×16=32 m/s$

During second stage, the height reached is,

$H2=v1t2+12a2t22=32×12+12×3×122=600 m$

At the end of second stage speed is,

$v2=v1+a2t2=32+3×12=68 m/s$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

$v3=v2−gt30=68−(9.8×t3)t3=689.8=6.938 sec$

The distance between the end of second stage and maximum altitude is,

$H3=v2t3−12gt32=68×6.938−12×9.81×(6.938)2=235.7 m$

So, the maximum height is,

$H1=H1+H2+H3=256+600+235.7=1091.7 m$

Answer 13

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Answer 14

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Answer 15

Expert Solution

Answer to Problem 82QAP

The average speed is $44.69 m/s$ and average velocity is 0.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The maximum altitude is $1091.67 m$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

$H=v3t4+12×g×t421091.7 m=0×t4+12×9.81×t42t4=14.91 sec$

Total time duration of flight,

$t=t1+t2+t3+t4=15+12+6.938+14.91=48.85 sec$

Total distance travelled is,

$Ht=H+H=2H=2×1091.7 m=2183.4 m$

So average speed will be,

$S=2183.3448.85=44.69 m/s$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

$v=Dit=0t=0 m/s$

Answer 20

Ch. 2 - Prob. 1QAP Ch. 2 - Prob. 2QAP Ch. 2 - Prob. 3QAP Ch. 2 - Prob. 4QAP Ch. 2 - Prob. 5QAP Ch. 2 - Prob. 6QAP Ch. 2 - Prob. 7QAP Ch. 2 - Prob. 8QAP Ch. 2 - Prob. 9QAP Ch. 2 - Prob. 10QAP

(a) The maximum altitude of the rocket.

Concept explainers

Videos

(a)

The maximum altitude of the rocket.

Answer to Problem 82QAP

Explanation of Solution

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Answer to Problem 82QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions

(a) The maximum altitude of the rocket.

Concept explainers

Videos

(a) The maximum altitude of the rocket.

Answer to Problem 82QAP

Explanation of Solution

(b) The average speed and average velocity when the rocket fall back to launch pad.

Answer to Problem 82QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions

(a)

The maximum altitude of the rocket.

(b)

The average speed and average velocity when the rocket fall back to launch pad.