What is the initial velocity of the rocket?

  • Thread starter suxatphysix
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In summary, the group of people watching the fireworks display witnessed an explosion 3.5 seconds after hearing a rocket launch. The explosion occurred 50.0 meters away from the launch point at an angle of 73.0 degrees above the horizontal. To determine the initial velocity of the rocket, the horizontal distance covered and the initial angle of launch must be calculated. Assuming the fuel is spent to launch the rocket, the initial speed can be calculated from the launching angle and horizontal speed. The rocket is accelerating under the influence of gravity, so equations for uniform acceleration in the negative y-direction must be used.
  • #1
suxatphysix
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Homework Statement


A group of people are watching a fireworks display. 3.5 s after they hear a rocket launch, they see the explosion. the explosion is observed to be located 50.0 from the launch point at an angle of 73.0 degrees above the horizontal.


(a) Sketch the problem. Label all parts of the problem with the appropriate variables. This includes all variables you will use in your equations. Specify the value of all the variables you are given.

(b) What is the initial velocity of the rocket?

(c) What is the velocity of the rocket just before it explodes?



Homework Equations



v = v0 + at
d=vt

The Attempt at a Solution



First I have a couple of questions.

Is the rocket accelerating? If yes, is it in the x and y directions?

Is there an assumption I should be making to solve this problem?

cos73 = x/50m
x= cos73(50m)
x=14.62

sin 73 = y/50m
y=sin73(50m)
y=47.82

v = v0 + at
0 = v0 + (-9.80)(3.5s)
vy0 = 34.3 m/s

d=vt
v=d/t
v=4.18 m/s

Then I would use the pythagorean theorem to solve for the final velocity but this is all done assuming that the velocity was constant and that the rocket was not accelerating.
 
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  • #2
bump bump ////
 
  • #3
suxatphysix said:

Homework Statement


A group of people are watching a fireworks display. 3.5 s after they hear a rocket launch, they see the explosion. the explosion is observed to be located 50.0 from the launch point at an angle of 73.0 degrees above the horizontal. (a) Sketch the problem. Label all parts of the problem with the appropriate variables. This includes all variables you will use in your equations. Specify the value of all the variables you are given.

(b) What is the initial velocity of the rocket?

(c) What is the velocity of the rocket just before it explodes?

Homework Equations



v = v0 + at
d=vt

The Attempt at a Solution



First I have a couple of questions.

Is the rocket accelerating? If yes, is it in the x and y directions?

Is there an assumption I should be making to solve this problem?

cos73 = x/50m
x= cos73(50m)
x=14.62

sin 73 = y/50m
y=sin73(50m)
y=47.82

v = v0 + at
0 = v0 + (-9.80)(3.5s)
vy0 = 34.3 m/s

d=vt
v=d/t
v=4.18 m/s

Then I would use the pythagorean theorem to solve for the final velocity but this is all done assuming that the velocity was constant and that the rocket was not accelerating.

the rocket is accelerating under the influence of gravity, so d=vt is not an equation you want to use. use the equations which are relevant for the case of uniform acceleration in the negative y-direction (at 9.8m/s^2) and no acceleration in the x-direction.
 
  • #4
If we assume that the fuel is spent to launch the rocket then we are dealing with parabolic motion.

Note that the initial speed, [tex]V[/tex], can be calculated from the launching angle, [tex]\theta _o[/tex], and the horizontal speed, [tex]V_x[/tex] (which can be calculated from the given data - I would think that the given distance is actually the horizontal distance covered up to the point of explosion)

[tex]V = \frac{V_x}{\cos(\theta _o)}[/tex]

to determine the initial angle use the known distances covered in the given time

[tex]x = Vt\cos(\theta _o)[/tex] ... (1)

and

[tex]y = Vt\sin(\theta _o) - \frac{1}{2}gt^2[/tex]

transferring the last term (which can be calculated) to the left gives

[tex]y + \frac{1}{2}gt^2 = Vt\sin(\theta _o)[/tex] ... (2)


dividing (2) by (1) now gives the launching angle [tex]\theta _o[/tex].
 
Last edited:

FAQ: What is the initial velocity of the rocket?

What is 2-D Kinematics?

2-D Kinematics is the study of motion in two dimensions, typically represented by the x and y axes. It involves analyzing the position, velocity, and acceleration of objects in two-dimensional space.

What is the difference between 1-D and 2-D Kinematics?

The main difference is the number of dimensions being considered. 1-D Kinematics only looks at motion in one dimension, while 2-D Kinematics adds another dimension, typically the y-axis. This allows for a more complex analysis of motion, as objects can move in both the horizontal and vertical directions.

How do you solve 2-D Kinematics problems?

To solve 2-D Kinematics problems, you need to break down the motion into its horizontal and vertical components. This can be done by using trigonometric functions and vector analysis. Then, you can apply equations of motion and use algebraic manipulation to solve for the desired variables.

What are some common examples of 2-D Kinematics in real life?

Some common examples include projectile motion, such as a ball being thrown or a cannon firing a cannonball, and circular motion, such as a car driving around a curved track. Other examples include the motion of objects in a rollercoaster or a pendulum swinging back and forth.

How is 2-D Kinematics used in other fields of science?

2-D Kinematics is used in many fields of science, such as physics, engineering, and biomechanics. It is used to analyze the motion of objects in various systems, including machines, structures, and living organisms. It is also used in computer graphics and animation to create realistic movements in virtual environments.

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