Asked 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?
Answered by drwls
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 = ?
Answered by Candy
Ok...explain to me how u obatined the final velocity
Answered by bobpursley
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.
Answered by Lindsay
But my number was correct, right? 17.714 m/s?
Answered by Lindsay
And drwls...where is 0.2041 coming from?
Answered by drwls
Your number was correct for Vy. You are asked to calculate Vx. You get it by knowing that Vx = sqrt(Vx^2 + Vy^2)
Answered by Lindsay
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?
Answered by Lindsay
All right n/m, I finally got the answer. :)
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