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Projectile Motion given starting height, initial angle, and range, find initial speed

  • Thread starter Potato21
  • Start date
  • #1
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Homework Statement


A person wants to shoot a rocket off of a mountain 3.26 km high so that the rocket, when shot at 31.3° above the horizontal from the very top of the mountain, lands at the foot of the mountain 9.37 km (horizontal distance) away. What initial speed must the rocket have? (Assume that the only force acting on the rocket after it is blasted from the top of the mountain is gravity.)


Homework Equations



df=1/2at^2+v0t+di

The Attempt at a Solution



I honestly don't know where to start, I've tried a variation of things and got nowhere :/

I'm trying to create formulas with vy and vx as vi sin 31.3 and vi cos 31.2 but I end up with the y side needing to use the quadratic formual with a vi variable in it and then I don't know how to solve it. Compared to the other questions given that also seems absurdly difficult. help?
 
Last edited:

Answers and Replies

  • #2
709
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firstly write down all the information the question gives you in component form, i.e. vertical,horizontal speed and acceleration, vertical horizontal displacements and also what you are required to find. Then find the relevant formula, your'e looking for s=ut +1/2at^2 with whatever symbols you use. The equation you gave seems to have one to many variables unless your teacher gave it to you for this specific case?
 
  • #3
10
0


firstly write down all the information the question gives you in component form, i.e. vertical,horizontal speed and acceleration, vertical horizontal displacements and also what you are required to find. Then find the relevant formula, your'e looking for s=ut +1/2at^2 with whatever symbols you use. The equation you gave seems to have one to many variables unless your teacher gave it to you for this specific case?
I wasn't given any equation :( its from my university's randomized homework question generator too so I can't ask any of my course-mates either. Their components end up as


y=-9.81/2t^2+visin31.3t+3250 and I'm looking for y=0m (converted from km)
x=vicos31.3t with x=9370m

I havent found a way to solve it though
 
  • #4
424
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see you know range(R) so
R = t * v * cos(31.3)

from here you get a relation between time and velocity.

just substitute t in terms of velocity in the equation

y = v *sin(31.1)*t + 1/2 * g * t^2

Here you know y and g, only v is unknown...
Solve!! :smile:
 
  • #5
10
0


see you know range(R) so
R = t * v * cos(31.3)

from here you get a relation between time and velocity.

just substitute t in terms of velocity in the equation

y = v *sin(31.1)*t + 1/2 * g * t^2

Here you know y and g, only v is unknown...
Solve!! :smile:
Does that range formula give show the range though when the landing position is not at the same height as the firing position?

EDIT:wait, yes it does, ALRIGHT! time to try this! I hope I don't have anything complicated in doing a quadratic formula though :S
 
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  • #6
709
0


Yep I agree completely, one interesting thing is that I got around 490m/s which is very fast but I suppose its a rocket and has a long way to go in 22.3 seconds.

Yous use crazy symbols though, what happened to conventional u,v and subscript i and j for relative direction.
 
  • #7
709
0


I believe that what yous are calling range is the horizontal displacement? and yes it can be thought of as independent of the vertical displacement but it is dependent on the time (which is dependent on vertical displacement).
 
  • #8
10
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Alright, I ended up with 42.7s and then solving for initial speed I got 257m/s which the computer accepts as right. Thank you everyone! I was trying to solve for initial velocity in one go which made me solve the range horizontal displacement formula for initial speed instead of time making it needlessy complicated. The next question was what asked for the time though so I assumed I needed to find the initial speed first.
 
  • #9
709
0


257m/s? wonder what I stuffed up.
 
  • #10
rl.bhat
Homework Helper
4,433
7


Alright, I ended up with 42.7s and then solving for initial speed I got 257m/s which the computer accepts as right. Thank you everyone! I was trying to solve for initial velocity in one go which made me solve the range horizontal displacement formula for initial speed instead of time making it needlessy complicated. The next question was what asked for the time though so I assumed I needed to find the initial speed first.
You can get the initial velocity in one go by using the equation

-y = v*sinθ*(x/v*cosθ) - 1/2*g*x^2/(v^2*cos^2θ)

-y = x*sinθ - 1/2*g*x^2/(v^2*cos^2θ)

Now solve for v.
 

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