Question on Projectile Motion equation [closed]

Question

A golf ball is shot into the air from the ground. If the initial horizontal velocity is 20m/s and the initial vertical velocity is 30m/s, what is the horizontal distance the ball will travel before it hits the ground?

Ans: The golf ball will continue to travel through air until gravity brings it back down to the ground. Under the gravity force, the vertical location of the ball as a function of time is given by the following equation:- $$ z = z_0 + v_{z_0} t - \frac{1}{2} g t^2 $$

The time it takes the golf ball to return the ground is equal to $$ T = \frac{2v_{z_0}}{g} $$

The horizontal distance the ball will travel before it hits the ground can be computed using the following equation:- $$ \Delta x = x-x_0 = v_{x_0} t = 2 v_{x_0} v_{z_0} / g = 122.4\text{ m} $$

The problems are:

1) what is the parameter $v$ represent?

2) In the example above $ \Delta x = x-x_0 = v_{x_0} t = 2 v_{x_0} v_{z_0} / g = 122.4\text{ m} $ What is the actual value in this equation? So that $x=?$ $x_0=?$ $v=30?$ $g=10?$ $z0=?$

Answer 1

I think your formatting is messing things up. There are two v's, $v_{x0}$ (i.e horizontal velocity) and $v_{z0}$ (i.e. vertical velocity at $t = 0$). You can get these values right out of the problem statement. Also, $g$ is 9.8m/s. Lastly, $x$ is insignificant; you only need to find $\Delta x$ (even if you were given $x$ it wouldn't effect your final answer).