• Support PF! Buy your school textbooks, materials and every day products Here!

2D Projectile Motion

  • Thread starter firefly_1
  • Start date
  • #1
16
0
[SOLVED] 2D Projectile Motion

So, this should be a relatively easy problem but I am thoroughly stumped and haven't a clue where to start.

A mortar used to launch fireworks fires a shell with an initial velocity [tex]v_{i}=130\mbox{m/s}[/tex] at [tex]\theta_{i}=60^o[/tex] above the horizontal. The shell explodes directly over a safety officer d = 850 meters downrange from the mortar. Assuming the ground is level, how high above the ground is the shell when it explodes? Is the shell falling as shown in the figure (refer to attachment) or is the shell still rising when it explodes? Assume that only the force of gravity acts on the shell as it flies.

Given information:
[tex]v_{i}=130[/tex] m/s
[tex]\theta_{i}=60^o[/tex]
d = 850 m
[tex]\Delta y_{i} =0[/tex] m

Any help that can be given would be greatly appreciated to at least get me started in the right area.
 

Attachments

Answers and Replies

  • #2
radou
Homework Helper
3,115
6
What are the basic kinematic equations you know?
 
  • #3
16
0
I know the 5 major ones.

[tex]a_{x} = constant[/tex]
[tex]v_{fx} - v_{ix} = a_{x}\Delta t[/tex]
[tex]\Delta x = (\frac{v_{ix} + v_{fx}}{2})\Delta t[/tex]
[tex]\Delta x = v_{ix}\Delta t + \frac{1}{2}a_{x}\Delta t^{2}[/tex]
[tex]v_{fx}^{2} - v_{ix}^{2} = 2a_{x}\Delta x[/tex]

and then also how to convert them to work with a free fall problem and function with gravity and [tex]\Delta y[/tex]
 
  • #4
16
0
I think I have the right answers now but if you could check over my work and see, it would be appreciated.

Given:
[tex]\theta = 60^{o}[/tex]
[tex]\Delta x = 850m [/tex]
[tex]g = 9.8 m/s^{2}[/tex]
[tex]v_{i} = 130 m/s [/tex]

Find: [tex]\Delta Y_{f}[/tex] and whether or not the projectile is rising or falling

Solution: Here's the RIGHT equations (and is the reason I was having such an issue to start with)

[tex]a_{x} = 0[/tex]
[tex]v_{fx} = v_{i}cos\theta[/tex]
[tex]\Delta x = (v_{i}cos\theta)\Delta t[/tex]
[tex]a_{y} = -g[/tex]
[tex]v_{fy} = v_{i}sin\theta - g\Delta t[/tex]
[tex]\Delta y = (v_{i}sin\theta)\Delta t - \frac{1}{2}g\Delta t^{2}[/tex]

Here is my actual work:
[tex]\Delta x = (v_{i}cos\theta)\Delta t[/tex]
[tex]850 = ((130)cos(60))\Delta t[/tex]
[tex]\frac{850}{(130)cos(60)} = \Delta t[/tex]
[tex]\Delta t = \mbox{13.08s}[/tex]

[tex]\Delta y = (v_{i}sin\theta)\Delta t - \frac{1}{2}g\Delta t^{2}[/tex]
[tex]\Delta y = ((130)sin(60))(13.08) - \frac{1}{2}(9.8)(13.08^{2}[/tex]
[tex]\Delta y = \mbox{634.266m}[/tex]

[tex]v_{fy} = v_{i}sin\theta - g\Delta t[/tex]
[tex]v_{fy} = (130)sin(60) - (9.80)(13.08)[/tex]
[tex]v_{fy} = \mbox{-15.6 m/s or 15.6 m/s down}[/tex]

So the projectile is 634.27m off the ground when it explodes and was falling.
 

Related Threads on 2D Projectile Motion

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
852
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
822
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
13
Views
767
Top