Projectile, Finding Initial Velocity

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Homework Help Overview

The problem involves a rocket launched from a height of 3220 meters at an angle of 35.3 degrees to the horizontal, with the goal of determining its initial velocity based on the horizontal distance it travels, which is 9580 meters.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to rearrange kinematic equations to express time and initial velocity in terms of each other. Some participants question the use of signs in the equations and how the height of the mountain is incorporated into the calculations.

Discussion Status

Participants are actively discussing the equations used and the assumptions made during the calculations. There is recognition of potential errors in the simplification steps, and some guidance has been offered regarding the treatment of vertical and horizontal components of motion.

Contextual Notes

There are indications of confusion regarding the correct application of kinematic equations and the treatment of the rocket's trajectory, particularly in relation to the height from which it is launched.

Laura1321412
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Homework Statement


A rocket is shot off a 3220 M high mountian at an angle of 35.3 to the horizontal. The rocket lands 9580 m away from the mountian, what was the initial velocity?


Homework Equations



y=Vi(y) *t +1/2gt^2

x= t*Vi(x)

Vi(y)= sin 35.3 Vi
Vi(x)= cos 35.3 Vi

The Attempt at a Solution



I rearanged the top equations to solve for t, then equated them, and replaced subbed the variables so the only var was Vi. When solving for Vi, i got some weird equations i don't know if I am using the right equations above, or what.

answers I've gotten: 318.7, 251.74

but neither are correct.

I don't know what to do! :(
 
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Take care of the signs. If the vertical component of the initial velocity is positive the acceleration has to be negative.
How did you use the height of the mountain? ehild
 
Here exactly what i did,,, sorry i should have written the whole thing out above.

time till rocket reaches highest point = Vi(y)/9.8
So, distance from top of mountian to highest point = Vi(y)/2 * Vi(y)/ 9.8 = Vi(y)^2/19.6

Then, distance from ground to highest point = Vi(y)^2/19.6 +3220
so, Vi(y)^2/19.6 +3220= Vi(0)*t + 1/2*9.8*t^2
- isolate t
sqrt(vi(y)^2/96.04 +657.14) = t time from highest point of flight to ground

So, total time for flight of rocket in y direction = sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8

And total flight time for x -> t = 9580/Vi(x)

Since the two t values have to be the same i equated the expressions

sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8 = 9580/Vi(x)

simplified a bit

vi(y)^2/96.04 +657.14 +vi^2/96.04 = 9580^2/Vi(x)^2

2Vi(x)^2Vi(y)^2/96.04 +657.14Vi(x)^2 = 9580^2

So, Vi(x)= cos 35.3Vi
Vi(y) = Sin 35.3 Vi

Sub these in

2(cos35.3Vi)^2(sin35.3Vi)^2/96.04 + 657.14(cos35.3Vi)^2= 9580^2

0.004632Vi^4 + 437.71Vi^2 = 9580^2

.0680Vi^2 + 20.92 Vi = 9580

Vi = 251.82


Checking this back with the values, this is too slow...

Im really confused! :S
 
Laura1321412 said:
So, total time for flight of rocket in y direction = sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8

And total flight time for x -> t = 9580/Vi(x)

Since the two t values have to be the same i equated the expressions

sqrt(vi(y)^2/96.04 +657.14) + Vi(y)/9.8 = 9580/Vi(x)

It is correct up to here, but the following is wrong.

Laura1321412 said:
simplified a bit

vi(y)^2/96.04 +657.14 +vi^2/96.04 = 9580^2/Vi(x)^2


(a+b)^2 is not a^2+b^2


ehild
 

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