Projectile Motion

  • Thread starter Physik
  • Start date
  • #1
24
0
Hello. I've been fustrated all day because I couldn't figure this problem out. I know, it may be simple, but trust me, I almost ripped my hair out trying to solve this problem. I would have been able to figure this problem out last year in basic physics, but I have forgotten 75% of physics I learned last year. Any help at all would be great.

Problem: A garden hose lying on the ground shoots out water at 6.5 m/s. The water travels 2 m. Find the angle at which the hose must be for the water to reach 2 m.

http://img381.imageshack.us/img381/117/prob5va.png [Broken]
 
Last edited by a moderator:

Answers and Replies

  • #2
208
0
try using the equation for range of a projectile

R = (v^2 * sin (2 * @)) / g
 
  • #4
20
0
That's no way to help him!

You just need to remember that the key to these sorts of problems is Time.

Decompose the velocity:
v(hor)=v*[email protected]
v(vert)=v*[email protected]

The horizontal velocity will never change, as there is no horizontal acceleration (assuming friction is barred). So the time the object should stay in the air is:

d=v*[email protected]*t,
t=d/(v*[email protected])

Now how long does the object stay in the air? It has a 'parabolic' path, so it's final vertical velocity is the negative of its initial. There is a constant acceleration, gravity, so the time it spends in the air is:

a*t = v(final) - v(initial) [now, as said, the v(final) is just negative v(initial)]
g*t = - 2 v(initial) [where g = -9.81 m/s^2]
[It is important to remember that the 'v(initial) here is the vertical velocity, which was v*[email protected]]
g*t = -2*v*[email protected]
t = -2*v*[email protected]/g

Now we have 2 expressions for t, put them together:
d/(v*[email protected]) = -2*v*[email protected]/g
d = -v^2*(2*[email protected]*[email protected])/g [double-angle identity: [email protected]@[email protected]]
d = -v^2*[email protected]/g, again where g=-9.81

That's where mathmike's formula came from. Now just plug in.
 
  • #5
346
0
Well, thats no way to help him either...

When the water comes out of the hose at 6.5m/s, there is a horizontal component to its velocity and a vertical component to its velocity.

The horizontal velocity is given by: [tex] v_h=6.5cos\theta[/tex],

while the vertical velocity is given by: [tex] v_v=6.5son\theta [/tex]

Remember, the duration of the projectile in motion is determined by the vertical velocity. The horizontal velocity is always constant assuming no viscous forces act on the projectile.

the time of flight is given by:
[tex] a=\frac{v-u}{T} [/tex]
[tex] T = 2(\frac{0-6.5cos\theta}{-9.81}) [/tex]

By equating the time of flight with the distance travelled by the projectile which can be found by dividing the distance moved by the projectile over its horizontal velocity, you can solve for [tex] \theta [/tex]
 

Related Threads on Projectile Motion

  • Last Post
Replies
11
Views
780
  • Last Post
Replies
1
Views
729
  • Last Post
Replies
2
Views
446
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
872
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
846
  • Last Post
Replies
1
Views
1K
S
Top