Projectile motion and initial speed

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SUMMARY

The problem involves calculating the initial speed of a ball thrown horizontally from a height of 23 meters, which hits the ground with a final speed three times its initial speed. The vertical acceleration is -9.8 m/s², and the relevant equations of motion include V = V0 + at and x - x0 = V0t + 0.5at². The time of flight was calculated to be approximately 2.1447 seconds, which is essential for determining the final velocity in the vertical direction. The solution requires applying vector algebra to relate the final and initial velocities.

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Homework Statement



From a height of 23 m, a ball is thrown horizontally. If it hits the ground with a speed that is three times its initial speed, what is the initial speed?

vertical: a = -9.8, x = -23, t = ?, V0=?, V = 3V0
horizontal: ?
x is displacement

Homework Equations


Possibly relevant:
V = Vo + at
x-x0 = volt + .5at2
x-x0 = Vt - .5at2
x-x0 = .5(V0+ v)t
V2 = V02 + 2a(x-x0)

The Attempt at a Solution


Seems simple, but I have tried and cannot do this problem. Seems like there is not enough info.

Solving the system of the following equations for vertical motion
x-x0 = volt + .5at2
x-x0 = Vt - .5at2
found that t = 2.1447, but I am not sure if this is correct. After that I just didn't know what to do. For horizontal movement, it seems like there is only time and the fact that the final velocity is 3 times the initial velocity.
 
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Your time for t is basically correct, though the number I got is slightly off. I get 2.17 for some reason.

Using t, you can also find the final velocity in the y direction.

After that, it's simple vector algebra. You know the Vfx vector and the Vfy vector. You know that the Vf vector is 3 times the Vox vector.

So, you have enough information. Just set up the vectors.
 

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