### Rocket Motion ProblemA rocket is launched at an angle of **51.0°** above the horizontal with an initial speed of **101 m/s**. The rocket moves for **3.00 s** along its initial line of motion with an acceleration of **32.0 m/s²**. At this time, its engines fail and the rocket proceeds to move as a projectile. #### Question (a): Maximum AltitudeFind the maximum altitude reached by the rocket.- Calculated Value: **1543.28 m** (Correct Answer) #### Question (b): Total Time of FlightFind its total time of flight.- Calculated Value: **34.2 s** (Incorrect Answer)#### Question (c): Horizontal RangeFind its horizontal range.- Calculated Value: **3873.6 m** (Incorrect Answer)### ExplanationThis problem involves the motion of a rocket launched at an angle with an initial velocity, followed by uniform acceleration for a short duration and then moving as a projectile after engine failure. For part (a), the correct maximum altitude reached by the rocket was calculated as 1543.28 meters.For parts (b) and (c), the values given—34.2 seconds for total time of flight and 3873.6 meters for horizontal range—were marked as incorrect answers. To fully solve these, one would need to break down the motion into two phases: powered flight with continuous acceleration and projectile motion after the engine failure. ### Detailed Calculation Steps:#### **Maximum Altitude**:- Combine vertical motion equations considering the powered phase until engine failure and subsequent projectile motion.#### **Total Time of Flight**:- Find the time for ascent using kinematic equations, then calculate the descent time using standard projectile motion equations.#### **Horizontal Range**:- Use the horizontal component of initial velocity and total time of flight to determine the horizontal displacement.Ensuring the right steps and calculations can help find the correct solutions for total time of flight and horizontal range.

given that, The angle of the launch of the rocket is: α=51. The initial speed of the rocket is:…

A rocket is launched at an angle of 51.0° above the horizontal with an initial speed of 101 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 32.0 m/s². At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. 1543.28 m (b) Find its total time of flight. 34.2 X S (c) Find its horizontal range. 3873.6 Xm

A rocket is launched at an angle of 51.0° above the horizontal with an initial speed of 101 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 32.0 m/s². At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. 1543.28 m (b) Find its total time of flight. 34.2 X S (c) Find its horizontal range. 3873.6 Xm

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A stone is thrown vertically upward with an initial speed of 20 m/s. When it returns, it fall into a…

A: The initial velocity of the stone, v = 20 m/s The initial position of the stone can be considered…

Q: The vector position of a particle varies in time according to the expression 7(t)=(1.76 t + 1.56)7…

A: Given data: The equation of particle which varies in time is given as, The equation of velocity…

Q: A projectile is launched in the air from the ground with an initial velocity v0=66 m/s. What the…

A: Maximum height is given by, hmax=v20sin2θ2g

Q: A projectile is launched with an initial speed of 637 m/s. Additionally, Viy = 588 m/s & Vix = -245…

A: Given V = 637 m/s Viy = 588 m/s Vix = -245 m/s We have to calculate the value of Vx = ? Vy = ? ax =…

Q: The velocity of a particle is given by v = {16t²i + 4t³j + (5t +2)k}m/s, where t is in seconds. If…

Q: A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the…

A: The initial velocity of the rocket is 12.0 m/s. The height of the building is 30.0 m. The horizontal…

Q: Magnitude and direction of the position. magnitude direction Magnitude and direction of…

Q: 2.59 A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s…

Q: A ball is thrown straight up into the air with an initial velocity of 13.0 m/s. Take the point of…

Q: The figure above is for a projectile on another planet where g is not 10 m/s2. Ignore air…

Q: for 5.00 seconds. After this 5.00-second interval, the object then travels as a projectile.…

A: Projectile A projectile is an object that is thrown without any fuel. Generally, a projectile is…

Q: A particle initially located at the origin has an acceleration of a - 2.00j m/s? and an initial…

Q: A particle is initially moving along the positive x-axis at a speed of 8.20 m/s. After 2.05 s, the…

A: As per given, Initial velocity u = 8.20 m/s Final velocity v = 4.67 m/s time taken by particle…

Q: On a spacecraft, two engines are turned on for 841 s at a moment when the velocity of the craft has…

Q: A mountain climber stands at the top of a 57.3 m cliff hanging over a calm pool of water. The…

Q: 1. A particle initially located at the origin has a constant acceleration of a = 3.00j m/s and an…

A: Given data The magnitude of the constant acceleration is a = 3 j m/s2 The initial velocity of the…

Q: A snowmobile is originally at the point with position vector 29.9 m at 95.0° counterclockwise from…

A: Step 1: here we have solved for question a Step 2: here we have solved for question b Step 3: don't…

Q: A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0s, thex…

A: The x-component of the initial velocity and acceleration of the puck is .The y-component of the…

Q: (a) Find the velocity and position vectors as functions of time t. m/s = m (b) Find the equation of…

A: a) The particle’s velocity can be determined as, The particle’s velocity can be determined as,

Q: A rock is thrown off a clift at an angle of 24.5 degrees above the horizontal. The clift in 168.9 m…

A: We have to know about motion of a projectile.

Q: A helicopter is ascending vertically with a speed of 5.21 m/sm/s . At a height of 102 mm above the…

Q: The acceleration of a particle is a = (5.00 i - 7.00 j) m/s^2. At t = 0, the particle leaves the…

Q: Suppose the position vector for a particle is given as a function of time by r(t) = x(t)î + y(t)ĵ,…

Q: A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0s, the x…

A: Write a given values of this question. v0x=+1.4m/sax=+7.5m/s2v0y=+9.2m/say=-6.4m/s2t=0.50s

Q: A missile silo is used to launch test rockets vertically upward out of the silo, giving the rocket…

Q: A rocket is launched at an angle of 59.0° above the horizontal with an initial speed of 101 m/s. The…

Q: A football is kicked from ground level with an initial velocity of 23.6 m/s at angle of 53.0° above…

Q: A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0s, the x…

Q: Use the given acceleration function and initial conditions to find the velocity vector v(t), and…

Q: A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0s, the x…

Q: A fireworks rocket explodes at a height of 42.0 m , producing fragments with velocities ranging from…

A: Initial velocity, Final velocity, Height, s = 42 m

Concept explainers

Relative Velocity

It is a measurement of an object with regards to the other object which is moving or stable or stationary. For example, if we are traveling in a bus and the bus is moving at speed of 50km/hr, then our speed is according to the nearby passenger on the bus is zero. According to them, we are not moving. But if one person standing on the ground or road observes us from outside of the bus, according to him we are moving at a speed of 50km/hr and the bus is also moving at a speed of 50km/hr. Now, the motion is recognized by the observer, it depends on a frame of reference or observer. This is motion called relative motion.

Question

### Rocket Motion Problem

A rocket is launched at an angle of **51.0°** above the horizontal with an initial speed of **101 m/s**. The rocket moves for **3.00 s** along its initial line of motion with an acceleration of **32.0 m/s²**. At this time, its engines fail and the rocket proceeds to move as a projectile.

#### Question (a): Maximum Altitude
Find the maximum altitude reached by the rocket.
- Calculated Value: **1543.28 m** (Correct Answer)

#### Question (b): Total Time of Flight
Find its total time of flight.
- Calculated Value: **34.2 s** (Incorrect Answer)

#### Question (c): Horizontal Range
Find its horizontal range.
- Calculated Value: **3873.6 m** (Incorrect Answer)

### Explanation
This problem involves the motion of a rocket launched at an angle with an initial velocity, followed by uniform acceleration for a short duration and then moving as a projectile after engine failure.

For part (a), the correct maximum altitude reached by the rocket was calculated as 1543.28 meters.

For parts (b) and (c), the values given—34.2 seconds for total time of flight and 3873.6 meters for horizontal range—were marked as incorrect answers. To fully solve these, one would need to break down the motion into two phases: powered flight with continuous acceleration and projectile motion after the engine failure.

### Detailed Calculation Steps:
#### **Maximum Altitude**:
- Combine vertical motion equations considering the powered phase until engine failure and subsequent projectile motion.

#### **Total Time of Flight**:
- Find the time for ascent using kinematic equations, then calculate the descent time using standard projectile motion equations.

#### **Horizontal Range**:
- Use the horizontal component of initial velocity and total time of flight to determine the horizontal displacement.

Ensuring the right steps and calculations can help find the correct solutions for total time of flight and horizontal range.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

The vector position of a particle varies in time according to the expression F(t)=(1.41 t + 1.72)i +(0.224t2 + 1.11)J where r is in meters and t is in seconds. Find the speed of the particle in m/s at 2.58 s.
A particle initially located at the origin has an acceleration of a = 1.00j m/s? and an initial velocity of v = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( 6 tî+ +5 t? j) m (b) Find the velocity of the particle at any time t. (6 î + 1 v tj) m/s (c) Find the coordinates of the particle at t = 6.00 s. X = 36 m y = 0 For the y coordinate of the particle, you need to know the y components of the initial position, the velocity, and the acceleration. m (d) Find the speed of the particle at t = 6.00 s. 8.49 m/s
An object has an initial velocity of 29.0 m/s at 95.0° and an acceleration of 1.90 m/s2 at 200.0°. Assume that all angles are measured with respect to the positive x-axis. (a) Write the initial velocity vector and the acceleration vector in unit vector notation. (b) If the object maintains this acceleration for 12.0 seconds, determine the average velocity vector over the time interval. Express your answer in your unit vector notation.
A particle initially located at the origin has an acceleration of a = 4.00j m/s? and an initial velocity of v = 9.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). tî + 2 j) m (b) Find the velocity of the particle at any time t. î + tj) m/s (c) Find the coordinates of the particle at t = 6.00 s. y%3D m (d) Find the speed of the particle at t = 6.00 s. m/s Need Help? Read It Watch It P Type here to search
A snowmobile is originally at the point with position vector 30.5 m at 95.0° counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with a constant acceleration 1.72 m/s2 at 200°. After 5.00 s have elapsed, find the following. (Express your answers in vector form.) (a) its velocity vector. (b) its position vector.
A snowmobile is originally at the point with position vector 27.1 m at 95.0° counterclockwise from the x axis, moving with velocity 4.13 m/s at 40.0°. It moves with constant acceleration 15 m at 200 After 5.00 s have elapsed, find the following. (Express your answers in vector form.) (a) its velocity vector (b) its position vector
PLEASE ANSWER ASAP: A 4.61m-tall truck is moving an initial speed of 15.32m/s due West with an acceleration of 0.22 m/s^2 with a 1.66m- person standing at the top of a truck. After 17.93s, the man throws a ball to the ground at shoulder level perpendicular to the left of the truck's direction with an initial velocity of 23.53 m/s at an angle of 42.80° above the horizontal. Assuming shoulder level is 80% of the total height and negligible air resistance, determine the magnitude and direction of the displacement of the ball upon impact. [Use positive angle and CCW]