You drop a water balloon on your friend’s head from

the third-story window of your apartment building,
15.0 m from the ground.

What is the water balloon’s acceleration while it is
in the air?
• How much time does it take for the water balloon
to fall?
• How fast is it going as it hits your friend’s head?
• How would these answers be different if, instead
of dropping the water balloon, you tossed it
straight upward with a speed of 5 m/s?

To answer these questions, we need to apply the principles of kinematics and gravity. Let's break it down step by step.

1. What is the water balloon's acceleration while it is in the air?
The acceleration of an object in free fall near the surface of the Earth is approximately equal to the acceleration due to gravity, denoted as "g." The value of g is approximately 9.8 m/s^2. Therefore, the water balloon's acceleration while falling will be 9.8 m/s^2, directed downwards.

2. How much time does it take for the water balloon to fall?
The time it takes for an object to fall from a certain height can be calculated using the kinematic equation:

y = (1/2)gt^2

where:
y = vertical height (15.0 m in this case)
g = acceleration due to gravity (9.8 m/s^2)

Rearranging the equation to solve for time (t), we get:

t = sqrt(2y/g)

Substituting the values, we have:

t = sqrt(2 * 15.0 m / 9.8 m/s^2) ≈ 1.92 seconds

So it would take approximately 1.92 seconds for the water balloon to fall.

3. How fast is it going as it hits your friend's head?
The final velocity of the water balloon just before it hits the ground can be found using another kinematic equation:

v = gt

where:
v = final velocity
g = acceleration due to gravity (9.8 m/s^2)
t = time of fall (1.92 seconds)

Substituting the values, we have:

v = 9.8 m/s^2 * 1.92 s ≈ 18.82 m/s

So the water balloon would be going approximately 18.82 m/s just before it hits your friend's head.

4. How would these answers be different if you tossed the water balloon straight upward with a speed of 5 m/s?
If you tossed the water balloon straight upward with an initial velocity of 5 m/s, the acceleration due to gravity would still be the same (9.8 m/s^2), but its direction would now be in the opposite direction (upwards).

In this case, the water balloon would eventually reach its maximum height and start falling back down. During this ascent, the balloon's velocity would decrease due to gravity until it reaches its highest point and starts descending. At the highest point, the velocity becomes zero before increasing downwards.

The time the balloon takes to reach its highest point can be calculated using the formula:

t = v / g

where:
t = time taken to reach the highest point
v = initial vertical velocity (5 m/s)
g = (-9.8) m/s^2 (negative since it is in the opposite direction)

Substituting values, we have:

t = 5 m/s / (-9.8 m/s^2) ≈ -0.51 seconds

Note that the time is negative because we are looking at the negative direction (upwards).

The time for the whole upward and downward journey would be twice this time (since total time = time up + time down), so:

Total time = 2 * 0.51 s ≈ 1.02 seconds

As for the final velocity when the balloon hits your friend's head, it would be the same as the initial upward velocity but in the opposite direction (downwards). So the final velocity would be -5 m/s.

Please note that these calculations assume a frictionless environment and neglect the air resistance. In reality, air resistance would impact the values slightly.