A) A hot air balloon has just lifted off and is rising at the constant rate of 1.70 m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 10.4 m/s. If the passenger is 3.00 m above her friend when the camera is tossed, how much time does it take for the camera to reach her?

B)How high is the passenger when the camera reaches her?

A) To find the time it takes for the camera to reach the passenger, we can use the equation of motion for vertical motion:

h = h0 + v0t + (1/2)gt^2

Where:
h = final height (the height of the passenger)
h0 = initial height (the height of the friend)
v0 = initial velocity (the initial speed of the camera)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2, assuming no air resistance)

Since the hot air balloon is rising and the passenger is above her friend, the initial height (h0) is negative. The equation becomes:

h = -3.00 m + 10.4 m/s * t - (1/2) * 9.8 m/s^2 * t^2

We want to find the time when the camera reaches the passenger, so we set h = 0 and solve for t.

0 = -3.00 m + 10.4 m/s * t - (1/2) * 9.8 m/s^2 * t^2

Rearranging the equation gives us a quadratic equation:

(1/2) * 9.8 m/s^2 * t^2 - 10.4 m/s * t + 3.00 m = 0

Now we can solve this quadratic equation using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

For this equation, a = (1/2) * 9.8 m/s^2, b = -10.4 m/s, and c = 3.00 m.

t = [-(10.4 m/s) ± sqrt((-10.4 m/s)^2 - 4 * (1/2) * 9.8 m/s^2 * 3.00 m)] / [2 * (1/2) * 9.8 m/s^2]

Simplifying further gives us:

t = [-(-10.4 m/s) ± sqrt((10.4 m/s)^2 - 2 * 9.8 m/s^2 * 3.00 m)] / [9.8 m/s^2]

t = [10.4 m/s ± sqrt(10.4^2 - 58.8)] / [9.8 m/s^2]

Using the quadratic formula, we can now calculate the value of t. We disregard the negative solution since time cannot be negative in this context.

t = [10.4 m/s + sqrt(10.4^2 - 2 * 9.8 m/s^2 * 3.00 m)] / [9.8 m/s^2]

Calculating this gives:

t ≈ 1.36 s (rounded to two decimal places)

Therefore, it takes approximately 1.36 seconds for the camera to reach the passenger.

B) To find the height of the passenger when the camera reaches her, we can substitute the value of t into the equation of motion:

h = -3.00 m + 10.4 m/s * t - (1/2) * 9.8 m/s^2 * t^2

Using the value of t we found earlier, we can calculate h:

h = -3.00 m + 10.4 m/s * 1.36 s - (1/2) * 9.8 m/s^2 * (1.36 s)^2

Simplifying this expression gives us:

h ≈ 7.17 m (rounded to two decimal places)

Therefore, the passenger is approximately 7.17 meters above the ground when the camera reaches her.

To find the time it takes for the camera to reach the passenger, we can use the equation of motion.

Step 1: Determine the relative velocity between the camera and the passenger. Since the hot air balloon is rising at a constant rate of 1.70 m/s and the camera is tossed straight upward with an initial velocity of 10.4 m/s, the relative velocity between the camera and the passenger is the difference of their velocities:

Relative velocity = initial velocity of the camera - ascending velocity of the balloon
Relative velocity = 10.4 m/s - 1.70 m/s
Relative velocity = 8.70 m/s

Step 2: Find the time it takes for the camera to reach the passenger. We can use the equation of motion:

Final position = Initial position + (Relative velocity * time)

In this case, the final position is the height of the passenger relative to the ground, which is 3.00 m. The initial position is 0 m (since the camera is thrown from the ground level). Let's assume time taken is "t":

Final position = Initial position + (Relative velocity * t)
3.00 m = 0 + (8.70 m/s * t)

Solving for t, we get:
t = 3.00 m / 8.70 m/s
t ≈ 0.34 seconds

Therefore, it takes approximately 0.34 seconds for the camera to reach the passenger.

To find the height of the passenger when the camera reaches her, we can use one of the equations of motion again.

Step 3: Calculate the height of the passenger when the camera reaches her. We can use the equation of motion:

Final position = Initial position + (Initial velocity * time) + (0.5 * acceleration * time^2)

In this case, the initial position is 3.00 m (the height of the passenger above her friend), the initial velocity is 0 m/s (since the passenger is stationary), and the time is 0.34 seconds (as calculated earlier). The acceleration is also 0 m/s^2 since the passenger's vertical velocity is constant.

Final position = 3.00 m + (0 m/s * 0.34 s) + (0.5 * 0 m/s^2 * (0.34 s)^2)

The middle term becomes zero because the initial velocity is zero. And the last term also equals zero since the acceleration is zero.

Final position = 3.00 m

Therefore, the height of the passenger when the camera reaches her is 3.00 meters.

camera:

Vi = 10.4

v = Vi - g t

h = 10.4 t - 4.9 t^2

passenger:
v = 1.7
h = 3 + 1.7 t

when are the two heights the same?

3 + 1.7 t = 10.4 t - 4.9 t^2
solve quadratic for t

then use h = 3 + 1.7 t to get height