A water balloon is thrown horizontally at a speed of 2.00 m/s from a roof of a building that is 6.00 m above ground. At the same instant the balloon is released a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much soon it hits the ground than the other balloon?

First Baloon

h = Vo*t + 0.5g*t^2 = 6 m.
0 + 4.9t^2 = 6
t^2 = 1.22
Tf = 1.11 s. = Fall time.

Second Baloon
2*t + 4.9t^2 = 6
4.9t^2 + 2t - 6 = 0.
Use Quadratic Formula:
Tf = 0.921 s. = Fall time.

1.11 - 0.921 = 0.189 s.
The Fall time of 2nd baloon was 0.189 s
shorter.

Well, that's an interesting situation we've got here! Let's break it down.

First, let's consider the horizontally thrown water balloon. Since there is no initial vertical velocity, the only force acting on it is gravity pulling it downward. However, since it's thrown horizontally, it will also experience horizontal motion. The vertical distance it needs to travel to hit the ground is 6.00 m.

Now, let's look at the balloon thrown straight down. It has an initial velocity of 2.00 m/s in the downward direction and it only needs to travel a vertical distance of 6.00 m to hit the ground.

Both balloons are experiencing the same amount of gravity, so they will both be accelerating downward at the same rate. As a result, the time it takes for an object to travel a certain vertical distance is only dependent on that distance and the acceleration due to gravity.

Using the equation of motion for free fall:

d = (1/2)gt^2

where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

We can rearrange this equation to solve for time:

t = sqrt(2d/g)

Let's calculate the time it takes for the horizontally thrown balloon to hit the ground:

t_horizontal = sqrt(2 * 6.00 m / 9.8 m/s^2) ≈ 1.09 s

Now, let's calculate the time it takes for the vertically thrown balloon to hit the ground:

t_vertical = sqrt(2 * 6.00 m / 9.8 m/s^2) ≈ 1.09 s

Huh, would you look at that! Both balloons hit the ground at the same time, approximately 1.09 seconds. So, neither balloon hits the ground sooner than the other.

Well, it seems like Mother Nature has played a little joke on us. No matter how the balloons are thrown, they both hit the ground at the same time. I hope that clears things up for you!

To solve this problem, we can use the equations of motion for both the horizontally thrown balloon and the vertically thrown balloon.

For the horizontally thrown balloon:
Initial vertical velocity (Vi) = 0 m/s (because it is thrown horizontally)
Horizontal distance (d) = the horizontal distance of the building, so d = 0 m
Vertical distance (h) = the height of the building, so h = 6.00 m
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)

Using the equation for vertical displacement:

h = Vi * t + (1/2) * g * t^2

Where t is the time taken to hit the ground.

Since the initial vertical velocity (Vi) is 0, the equation simplifies to:

h = (1/2) * g * t^2

Plugging in the values, we get:

6 = (1/2) * 9.8 * t^2

Simplifying further:

t^2 = (6 * 2) / 9.8
t^2 = 1.2245
t ≈ 1.107 s

Therefore, the vertically thrown balloon takes approximately 1.107 seconds to hit the ground.

For the vertically thrown balloon:
Initial vertical velocity (Vi) = -2.00 m/s (downwards)
Vertical distance (h) = 6.00 m
Acceleration due to gravity (g) = 9.8 m/s^2

Using the same equation as before for vertical displacement:

h = Vi * t + (1/2) * g * t^2

Since the initial vertical velocity (Vi) is -2.00 m/s, the equation becomes:

6 = -2.00 * t + (1/2) * 9.8 * t^2

Rearranging the equation:

4.9t^2 - 2t - 6 = 0

Solving this quadratic equation, we find:

t ≈ 1.665 s

Therefore, the vertically thrown balloon takes approximately 1.665 seconds to hit the ground.

Comparing the two times, we see that the horizontally thrown balloon hits the ground faster than the vertically thrown balloon. The horizontally thrown balloon hits the ground approximately 1.107 seconds after it is released, while the vertically thrown balloon hits the ground approximately 1.665 seconds after it is released.

Therefore, the horizontally thrown balloon hits the ground sooner than the vertically thrown balloon by approximately 1.665 - 1.107 = 0.558 seconds.

To determine which balloon hits the ground first, we can calculate the time it takes for each balloon to reach the ground. Let's start with the water balloon thrown horizontally.

1. Calculate the time taken for the horizontally thrown balloon to reach the ground:
The time it takes for the balloon to hit the ground horizontally does not depend on its initial velocity. It only depends on the height of the building. We can use the formula for free-fall motion:

distance = (1/2) * acceleration * time^2

In this case, the only acceleration acting on the balloon is due to gravity (9.8 m/s^2), and the distance is 6.00 m:

6.00 m = (1/2) * 9.8 m/s^2 * time^2

Simplifying the equation:

time^2 = (2 * 6.00 m) / 9.8 m/s^2
time^2 = 1.22 s^2

Taking the square root of both sides, we get:

time = √(1.22 s^2)
time ≈ 1.10 s

So, the horizontally thrown water balloon takes approximately 1.10 seconds to hit the ground.

2. Calculate the time taken for the vertically thrown balloon to reach the ground:
The time it takes for the vertically thrown balloon to hit the ground can also be calculated using free-fall motion. Since the balloon is thrown straight down, its initial velocity is directly downward, and we can use the same formula:

6.00 m = (1/2) * 9.8 m/s^2 * time^2

Simplifying the equation:

time^2 = (2 * 6.00 m) / 9.8 m/s^2
time^2 = 1.22 s^2

Taking the square root of both sides, we get:

time = √(1.22 s^2)
time ≈ 1.10 s

Just like the horizontally thrown balloon, the vertically thrown balloon also takes approximately 1.10 seconds to hit the ground.

Therefore, both balloons hit the ground at the same time, approximately 1.10 seconds after being released.