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Projectile, Finding Initial Velocity

  • #1

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 im using the right equations above, or what.

answers ive gotten: 318.7, 251.74

but neither are correct.

I don't know what to do! :(
 

Answers and Replies

  • #2
ehild
Homework Helper
15,478
1,854
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
 
  • #3
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
 
  • #4
ehild
Homework Helper
15,478
1,854
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.

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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