A stone thrown horizontally from a height of 7.8 m hits the ground at a distance of 12.0 m. Calculate the speed of the ball as it hits the ground.
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 4:40pm
The 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 4:46pm
15.6 m/s. I understand now, thanks a bunch.
Physics - Tom Bush, Sunday, February 8, 2009 at 8:06am
Yes, 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.