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.

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] =?

15.6 m/s. I understand now, thanks a bunch.

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.

To calculate the speed of the ball as it hits the ground, you can use the principles of projectile motion. Since the stone is thrown horizontally, its initial vertical velocity is zero. The only force acting on the stone is gravity, pulling it downward.

The distance traveled horizontally (12.0 m) does not affect the vertical motion of the stone. Therefore, you can focus on the vertical motion to find the final velocity.

The vertical motion of the stone can be described by the equation:

h = ut + (1/2)gt^2

Where:
- h is the height of the stone (7.8 m)
- u is the initial vertical velocity (0 m/s)
- g is the acceleration due to gravity (-9.8 m/s^2, assuming the stone is near the surface of the Earth)
- t is the time taken to reach the ground

Since the stone hits the ground, its final height (h) is 0. You can rearrange the equation to solve for t:

0 = (1/2)gt^2
0 = -4.9t^2

Solving for t, you find that t = 0, which is the initial condition. However, we are interested in the time taken to reach the ground, so we need to consider the positive root:

t = √(0/-4.9)
t = 0 s

Since the time taken to reach the ground is 0 seconds, you can conclude that the speed of the ball as it hits the ground is equal to the initial speed of the ball, which you found to be 9.52 m/s.