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September 30, 2014

September 30, 2014

Posted by **Lindsay** on Saturday, November 3, 2007 at 4:24pm.

I was able to find the initial speed of the ball, which is 9.52 m/s. However, I'm unsure about whether I need to use this number to get the speed as it hits the ground.

- Physics -
**drwls**, Saturday, November 3, 2007 at 4:40pmThe speed which which it was thrown is 12.0 m/(fall time) = The fall time is

T = sqrt (2H/g) = 1.262 s

Vox = 12.0/1.262 = 9.51 m/s. Close enough. You WILL need that number to finish the problem. It remains the horizontal component of velocity until it hits the ground.

The vertical component of velocity when it hits the ground is

Vyfinal = sqrt (2gH) = 12.36 m/s

When it hits the ground, the speed is

Vfinal = sqrt[Vox^2 + Vyfinal^2] =?

- Physics -
**Lindsay**, Saturday, November 3, 2007 at 4:46pm15.6 m/s. I understand now, thanks a bunch.

- Physics -
**Tom Bush**, Sunday, February 8, 2009 at 8:06amYes, that's great explain. Also, you be able to use suvat formulae in steps to work out time and then vox and then resultant velocity.

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