What is the initial velocity of the rocket?

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The discussion focuses on calculating the initial velocity of a rocket observed during a fireworks display, where the explosion occurs 3.5 seconds after launch at a distance of 50.0 meters and an angle of 73.0 degrees. Participants clarify that the rocket is accelerating due to gravity, which affects the equations used for calculations. The horizontal and vertical components of the motion are determined using trigonometric functions, leading to the conclusion that the initial speed can be derived from the horizontal distance and launch angle. The equations for motion under uniform acceleration are emphasized, particularly for determining the rocket's trajectory. Overall, the conversation highlights the importance of correctly applying physics principles to solve the problem.
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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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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.
 
If we assume that the fuel is spent to launch the rocket then we are dealing with parabolic motion.

Note that the initial speed, V, can be calculated from the launching angle, \theta _o, and the horizontal speed, V_x (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)

V = \frac{V_x}{\cos(\theta _o)}

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

x = Vt\cos(\theta _o) ... (1)

and

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

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

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


dividing (2) by (1) now gives the launching angle \theta _o.
 
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