What is the Initial Velocity of a Ball Thrown off a Cliff?

[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 that's all i can get, i have the y axe vector for the initial velocity and i can't seem to figure out anything further...help needed! lol

 

[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.

 

[QuantumKing]

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

 

[Astronuc]

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.

 

[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...