A rocket is launched to travel vertically upward with a constant velocity of 20 m/s. After travelling for 35 seconds, the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. The height achieved by it is:

Question

Prepp Team · Accepted Answer

Understanding the Rocket Motion Problem

This problem describes a rocket launch scenario that occurs in two distinct phases:

An initial phase where the rocket moves vertically upward with a constant velocity.
A second phase where the fuel supply is cut off, and the rocket continues to move under the influence of gravity alone, acting as a free body.

To find the total height achieved, we need to calculate the height covered in each phase and add them together.

Phase 1: Constant Velocity Ascent

In the first phase, the rocket travels vertically upward with a constant velocity for a given duration. The distance covered during this phase can be calculated using the simple formula: distance = velocity $\times$ time.

Constant velocity, $v_1 = 20$ m/s
Time duration, $t_1 = 35$ seconds

The height covered in Phase 1, let's call it $h_1$, is:

\[ h_1 = v_1 \times t_1 \] \[ h_1 = 20 \, \text{m/s} \times 35 \, \text{s} \] \[ h_1 = 700 \, \text{m} \]

So, the rocket travels 700 meters in the first 35 seconds.

Phase 2: Free Body Motion After Fuel Cut-off

When the fuel supply is cut off, the rocket becomes a free body, meaning it is only under the influence of gravity. It will continue to move upward initially because of the velocity it had at the end of Phase 1, but its speed will decrease due to gravity until it reaches its maximum height. At the maximum height, its vertical velocity becomes zero.

For this phase, we can use the equations of motion under constant acceleration, where the acceleration is due to gravity ($g$). We will assume $g = 10$ m/s$^2$ for calculations, as is common in many introductory physics problems unless specified otherwise. Since gravity acts downwards and we are considering upward motion as positive, the acceleration is $a = -g = -10$ m/s$^2$.

Initial velocity for Phase 2, $u_2$ = Velocity at the end of Phase 1 = 20 m/s
Final velocity at the maximum height, $v_2$ = 0 m/s
Acceleration due to gravity, $a = -10$ m/s$^2$

We want to find the additional height gained in this phase, let's call it $h_2$. We can use the kinematic equation relating initial velocity, final velocity, acceleration, and displacement:

\[ v_2^2 = u_2^2 + 2ah_2 \]

Plugging in the values:

\[ (0 \, \text{m/s})^2 = (20 \, \text{m/s})^2 + 2(-10 \, \text{m/s}^2)h_2 \] \[ 0 = 400 \, \text{m}^2/\text{s}^2 - 20 \, \text{m/s}^2 h_2 \]

Now, we solve for $h_2$:

\[ 20 \, \text{m/s}^2 h_2 = 400 \, \text{m}^2/\text{s}^2 \] \[ h_2 = \frac{400 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} \] \[ h_2 = 20 \, \text{m} \]

So, the rocket travels an additional 20 meters upwards after the fuel is cut off before it starts falling back down.

Calculating Total Height Achieved by Rocket

The total height achieved by the rocket is the sum of the height covered in the constant velocity phase and the maximum additional height covered in the free body phase.

Height in Phase 1, $h_1 = 700$ m
Additional height in Phase 2, $h_2 = 20$ m

Total Height, $H = h_1 + h_2$

\[ H = 700 \, \text{m} + 20 \, \text{m} \] \[ H = 720 \, \text{m} \]

Therefore, the maximum height achieved by the rocket is 720 meters.

Summary of Steps for Calculating Rocket Height

Calculate the distance covered during the initial phase of constant velocity motion using distance = velocity $\times$ time.
Determine the velocity of the rocket at the moment the fuel is cut off. This is the initial velocity for the subsequent free body motion.
Using kinematic equations for motion under constant acceleration (gravity), calculate the additional height the rocket ascends until its vertical velocity becomes zero. Remember to use the acceleration due to gravity ($-g$) as the acceleration.
Add the height from Phase 1 and the additional height from Phase 2 to find the total maximum height achieved.

Phase	Description	Initial Velocity	Final Velocity	Acceleration	Time / Displacement	Calculation	Height Gained
Phase 1	Constant Velocity Ascent	20 m/s	20 m/s	0 m/s$^2$	35 s	$h_1 = v_1 \times t_1$	700 m
Phase 2	Free Body Ascent (to max height)	20 m/s	0 m/s	-10 m/s$^2$ (assumed $g=10$)	N/A (calculated displacement)	$v_2^2 = u_2^2 + 2ah_2$	20 m
Total	Total Height	-	-	-	-	$H = h_1 + h_2$	720 m

Revision Table: Key Physics Concepts

Concept	Formula / Description	Application in Problem
Constant Velocity Motion	Distance = Velocity $\times$ Time	Calculating height in Phase 1
Free Body Motion	Motion under the sole influence of gravity. Acceleration is constant ($-g$).	Describing rocket motion in Phase 2
Kinematic Equations	$v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$	Calculating additional height in Phase 2
Acceleration due to Gravity ($g$)	Approx. 9.8 m/s$^2$ or 10 m/s$^2$ near Earth's surface. Acts downwards.	Used as acceleration in Phase 2 ($-g$)
Maximum Height in Projectile Motion	Point where vertical velocity is 0. Can be calculated using kinematic equations.	Finding the peak of the rocket's trajectory in Phase 2.

Additional Information: Rocket Physics and Kinematics

This problem combines two fundamental types of motion often studied in kinematics: constant velocity motion and motion under constant acceleration (specifically, free fall). Understanding how to break down complex problems into simpler parts is crucial.

When an object is in free fall, its acceleration is always $g$ downwards, regardless of its velocity. This is a key principle. Even when the rocket is momentarily at its highest point, its velocity is zero, but its acceleration is still $g$ downwards.

The choice of $g = 10$ m/s$^2$ is often made to simplify calculations in physics problems. In real-world scenarios, $g$ varies slightly depending on location and altitude, and 9.8 m/s$^2$ is a more precise average value. However, for multiple-choice questions, using the value that leads to one of the options is usually intended.

For upward motion, if we take the upward direction as positive, the acceleration due to gravity must be taken as negative ($-g$) because it acts in the opposite direction (downwards). If we take downwards as positive, then upward initial velocity would be negative, and gravity would be positive ($+g$). The convention chosen affects the signs in the equations but should result in the same magnitude for displacement or distance.