Irritating vector problem

1. Sep 21, 2006

QuantumKing

Some dude on a cliff throws a ball with an unknown initial velocity. The ball is in motion for 2seconds before its instant velocity is mesured, being 24m/s [E 45 S]. Determine its initial velocity.

i got

g= -9,8
t= 2
instV= 24
iV=? Vx2= 24cos45 = 17m/s
Vy2= 24sin45 = 17m/s
9,8=(17 - Vy1)/2.....Vy2=2,6m/s

and thats all i can get, i have the y axe vector for the initial velocity and i cant seem to figure out anything further...help needed! lol

2. Sep 21, 2006

civil_dude

Sounds like a poorly worded question. If on a cliff, then what is E 45 S? Those are horizontal plane directions. I am confused also.

3. Sep 21, 2006

QuantumKing

well i am assuming they mean 45 degrees under the horizontal axe of the top of the cliff

4. Sep 21, 2006

Staff: Mentor

I agree with civil_dude, the question is poorly worded, unless the problem has redefined the compass orientations. N, S, E, W normally refer to directions (lateral) normal to the local gravity field, which is vertical (elevation). E45S represents a vertical plane, in that orientation.

On the other hand,

if the velocity has no vertical component and is horizontal, pointing in that direction, then one has a horizontal speed of 24 m/s at 2 seconds, which would be at the top of an arc. So it would be 48 m away and would come back to same elevation 2 sec later for a range of 96 m.

Can one determine the altitude after 2 seconds, i.e. one needs to determine the angle the ball is thrown.

One may neglect air resistance.

5. Sep 21, 2006

QuantumKing

Its simply initial velocity components along with final velocity comps. and a vertical accelerated component. The final velocity, after 2 seconds, is 24m/s directed at a 45 degree angle under the horizontal pane of the top of the cliff. I found the vertical and horizontal final velocities, and managed to find the vertical initial velocity component, but am completely lost as to how to find the horizontal initial velocity comp...