A student throws a water balloon vertically downward from the top of a building.

The balloon leaves the thrower's hand with a speed of 6.00 m/s. Air resistance may be ignored,
so the water balloon is in free fall after it leaves the thrower's hand. (a) What is its speed after
falling for 2.00 s? (b) How far does it fall in 2.00 s?
(c) What is the magnitude of its velocity after falling 10.0 m?
(d) Sketch ay-t, vy-t, and y-t graphs for the motion.

The equations are (with distance Y and velocity V measured positive downward)

Y = 6t + 4.9 t^2
V = 6 + 9.8 t

Y=0 corresponds to the point at which it is dropped.

You do the graphing and plugging in of numbers, such as 2.00s for t in parts (a) and (b).

ay is 9.8 m/s^2 at all times it is falling

question one

(a) To determine the speed of the water balloon after falling for 2.00 s, we can use the formula for the final velocity in free fall:

v = u + gt,

where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time

Given:
u = 6.00 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
t = 2.00 s (time)

Substituting the given values into the formula, we have:

v = 6.00 m/s + (9.8 m/s^2)(2.00 s)
v = 6.00 m/s + 19.6 m/s
v = 25.6 m/s

Therefore, the speed of the water balloon after falling for 2.00 s is 25.6 m/s.

(b) To determine the distance the water balloon falls in 2.00 s, we can use the formula:

s = ut + (1/2)gt^2,

where:
s = distance
u = initial velocity
g = acceleration due to gravity
t = time

Given:
u = 6.00 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
t = 2.00 s (time)

Substituting the given values into the formula, we have:

s = (6.00 m/s)(2.00 s) + (1/2)(9.8 m/s^2)(2.00 s)^2
s = 12.00 m + (1/2)(9.8 m/s^2)(4.00 s^2)
s = 12.00 m + 19.6 m
s = 31.6 m

Therefore, the water balloon falls a distance of 31.6 meters in 2.00 s.

(c) To determine the magnitude of the velocity after falling 10.0 meters, we can use the formula for the final velocity in free fall:

v^2 = u^2 + 2gs,

where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
s = distance

Given:
u = 6.00 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
s = 10.0 m (distance)

Substituting the given values into the formula, we have:

v^2 = (6.00 m/s)^2 + 2(9.8 m/s^2)(10.0 m)
v^2 = 36.00 m^2/s^2 + 196.0 m^2/s^2
v^2 = 232.0 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 15.2 m/s

Therefore, the magnitude of the velocity after falling 10.0 m is approximately 15.2 m/s.

(d) To sketch the ay-t, vy-t, and y-t graphs for the motion, we need to consider the following information:

ay-t Graph:
The acceleration due to gravity, g, is constant (-9.8 m/s^2) as it acts downward. Therefore, the ay-t graph would be a horizontal line passing through -9.8 m/s^2.

vy-t Graph:
The initial velocity, u, is 6.00 m/s, and the acceleration due to gravity, g, is -9.8 m/s^2. The vy-t graph would start at 6.00 m/s and decrease linearly with time at a rate of 9.8 m/s^2.

y-t Graph:
The initial position, y0, is the height of the building, which is not provided in the given information. Assuming the initial position is 0 m, the y-t graph would be a quadratic curve (parabola) concave downward. The shape of the curve depends on the values of u, g, and any additional values given for the problem.

Please note that without the specific height of the building or any other reference points, the y-t graph cannot be accurately sketched.

To answer these questions, we can use the equations of motion for objects in free fall. Here's how you can solve each part:

(a) What is its speed after falling for 2.00 s?
The acceleration due to gravity is approximately 9.8 m/s^2. To find the speed, we use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given:
Initial velocity (u) = 6.00 m/s
Time (t) = 2.00 s
Acceleration (a) = 9.8 m/s^2 (downward)

Using the equation v = u + at:
v = 6.00 m/s + (9.8 m/s^2)(2.00 s)
v = 6.00 m/s + 19.6 m/s
v = 25.6 m/s

Therefore, the speed of the water balloon after falling for 2.00 s is 25.6 m/s.

(b) How far does it fall in 2.00 s?
To find the distance traveled (vertical displacement), we use the equation s = ut + 0.5at^2, where s is the displacement.

Given:
Initial velocity (u) = 6.00 m/s
Time (t) = 2.00 s
Acceleration (a) = 9.8 m/s^2 (downward)

Using the equation s = ut + 0.5at^2:
s = (6.00 m/s)(2.00 s) + 0.5(9.8 m/s^2)(2.00 s)^2
s = 12.0 m + 0.5(9.8 m/s^2)(4.00 s^2)
s = 12.0 m + 19.6 m
s = 31.6 m

Therefore, the water balloon falls 31.6 meters in 2.00 seconds.

(c) What is the magnitude of its velocity after falling 10.0 m?
To find the magnitude of the velocity, we can use the equation v^2 = u^2 + 2as.

Given:
Initial velocity (u) = 6.00 m/s
Displacement (s) = 10.0 m (downward)
Acceleration (a) = 9.8 m/s^2 (downward)

Using the equation v^2 = u^2 + 2as:
v^2 = (6.00 m/s)^2 + 2(9.8 m/s^2)(-10.0 m)
v^2 = 36.0 m^2/s^2 - 196.0 m^2/s^2
v^2 = -160.0 m^2/s^2 (negative value indicates downward direction)

Since velocity cannot be negative unless specified, we take the positive square root:

v = √(-160.0 m^2/s^2) = 12.65 m/s

Therefore, the magnitude of the velocity after falling 10.0 m is 12.65 m/s.

(d) Sketch ay-t, vy-t, and y-t graphs for the motion.
In text form, it is difficult to sketch graphs. However, I can describe each graph briefly:

- The ay (acceleration in the y-direction) versus t graph will be a horizontal line at -9.8 m/s^2, indicating constant acceleration due to gravity.
- The vy (velocity in the y-direction) versus t graph will start at 6.00 m/s and have a constant downward slope, since velocity increases with time due to acceleration.
- The y (vertical displacement) versus t graph will be a quadratic curve, with a concave downward shape, indicating a free-falling object.

Please note that the exact shapes of the graphs may vary depending on the chosen scale, but they would follow the descriptions given.

I hope this helps! Let me know if you have any further questions.