Physics
posted by Lindsay .
A ball is thrown horizontally from a height of 16.01 m and hits the ground with a speed that is 5.0 times its initial speed. What was the initial speed?
Ok so I got a vertical final velocity of 17.714 m/s. Now I need to find the horizontal speed, correct? Which equation can I use?

What you have calculated is the vertical component of the velocity at impact (Vy), not the full speed. The full speed at impact is sqrt (Vx^2 + Vy^2) You need Vx, which is the same as the horizontal velcoity when it was thrown. Vx was the initial speed.
sqrt (Vx^2 + Vy^2) = 5 Vx
Vx^2 + Vy^2 = 25 Vx^2
Vy = sqrt(24)* Vx
Vx = 0.2041 Vy = ? 
Ok...explain to me how u obatined the final velocity

The final velocity contains two perpendicular components, horizontal and vertical. Using the pythoreagan theorm,
vf= sqrt(vhoriztal^2+ vvertical*2)
In this case, one knows that Vf=4vhorizontal, and one knows the Vvertialfinal.
Solve for vhorizontal. 
But my number was correct, right? 17.714 m/s?

And drwls...where is 0.2041 coming from?

Your number was correct for Vy. You are asked to calculate Vx. You get it by knowing that Vx = sqrt(Vx^2 + Vy^2)

All right. So this is what my equation looks like so far. I need to solve for Vx:
25Vx^2 = 313.786 + Vx^2
Correct? 
All right n/m, I finally got the answer. :)
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