Projectile, Finding Initial Velocity

AI Thread Summary
The discussion revolves around calculating the initial velocity of a rocket launched from a height of 3220 meters at a 35.3-degree angle, landing 9580 meters away. Participants are attempting to apply projectile motion equations but are struggling with the correct application and simplification of terms. One user has derived various equations but is confused about the results and the signs in their calculations. A key point of contention is the proper handling of the time equations for both vertical and horizontal motion, with an emphasis on ensuring they equate correctly. The conversation highlights the importance of careful algebraic manipulation and the correct use of trigonometric components in projectile motion problems.
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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