You throw a ball straight into the air from a height of 4 feet and with a speed of 8.22 m/s. The moment the ball leaves your hand you start running away at a speed of 3.36 m/s. How far are you from the ball, the moment it hits the ground?

(what are the formulas used?)

here is the method:

http://www.jiskha.com/display.cgi?id=1391459618

how did you find t? my physics professor is not the greatest at explaining so I am extremely lost and don't know at all how to do this.

I solved the quadratic equation

if
a x^2 + b x + c = 0
then
x = [-b +/- sqrt (b^2-4ac) ]/2a

You had this in algebra I suspect.
It works fine for
0 = (Hi-h) + Vi t - (1/2) g t^2

as with any parabola, it crosses the finish height twice, once on the way up and again later on the way down. We want it later coming back down so use the + sqrt and not the - sqrt for t

okay, now I get it. thank you for your help!

To determine the distance between you and the ball when it hits the ground, we can break down the problem into two parts: the vertical motion of the ball and the horizontal motion of yourself.

First, let's calculate the time it takes for the ball to hit the ground. We can use the equation for vertical motion, specifically the equation for displacement in the y-direction:

y = y0 + v0y * t - 0.5 * g * t^2

where:
y = final displacement (in this case, y = 0 since the ground is the reference point)
y0 = initial displacement (4 feet, which is 1.22 meters)
v0y = initial vertical velocity (8.22 m/s, directed upwards)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken

Using the equation above, we can solve for t:
0 = 1.22 + 8.22 * t - 0.5 * 9.8 * t^2

This equation is a quadratic equation, which can be solved using various methods (e.g., factoring, quadratic formula, graphing). After solving the equation, we find that t ≈ 0.915 seconds.

Now let's find the horizontal distance you have traveled during 0.915 seconds. We can use the equation for displacement in the x-direction:

x = v0x * t

where:
x = horizontal displacement (this is the distance you're looking for)
v0x = initial horizontal velocity (3.36 m/s, which remains constant throughout the motion)
t = time taken (0.915 seconds)

By substituting the given values, we can calculate the horizontal displacement:
x = 3.36 m/s * 0.915 s

Evaluating this equation, we find that x ≈ 3.0756 meters.

Therefore, you would be approximately 3.0756 meters away from the ball horizontally when it hits the ground.