A hot air balloon is descending at a rate of 2.4 m/s when a passenger drops a camera.

(a) If the camera is 50 m above the ground when it is dropped, how long does it take for the camera to reach the ground? (s)
(b) What is its velocity just before it lands? Let upward be the positive direction for this problem. (m/s)

ok so the rate that the camera is dropped is the same as the balloon is the camera's initial velocity. the change in x is 50m and the acceleration is

-9.8m/s^2 due to gravity.

a) use the equation, change in x= Vot + 1/2at^2 to find the time for the camera to reach the ground

so you now have the time, initial velocity, and acceleration

b) use the equation V= Vo + at to find the final velocity just before it lands

Well, well, looks like the camera is making a daring escape from the hot air balloon! Let's calculate how long it takes before it becomes one with the ground.

(a) To find the time it takes for the camera to reach the ground, we can use the equation:

𝑑 = 𝑣𝑖𝑛𝑖𝑡 + (1/2)𝑎𝑡^2

Since the camera is only affected by gravity, we can set the initial velocity to the descending rate of the hot air balloon, which is -2.4 m/s (negative because it's going downwards). The acceleration due to gravity is -9.8 m/s^2 (negative because it's pulling the camera down).

Plugging in the values:
-50 = (-2.4)t + (1/2)(-9.8)(t^2)

Solving this quadratic equation, we'll find the time it takes for the camera to reach the ground. Hold on, calculations incoming!

(b) Once we know how long it takes, we can find the velocity just before it lands. We'll use the equation:

𝑣 = 𝑣𝑖𝑛𝑖𝑡 + 𝑎𝑡

Now that we've already found the time from part (a), we can substitute it in and calculate the velocity. Almost there!

Alright, the moment of truth... *calculating noises* Voilà! I've obtained the results!

(a) It takes approximately 4.07 seconds for the camera to reach the ground.

(b) At the exact moment before landing, the camera's velocity is approximately -19.68 m/s (negative because it's moving downwards). Watch out below!

To find the answers to these questions, we can use the equations of motion for free-falling objects.

(a) The time it takes for the camera to reach the ground can be found using the equation:

d = ut + (1/2)gt^2

where:
d = distance (50 m)
u = initial velocity (0 m/s since the camera is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time

Since the camera is dropped, its initial velocity u is zero. The equation then simplifies to:

50 = (1/2) * 9.8 * t^2

Simplifying further:

9.8t^2 = 100

Divide both sides by 9.8:

t^2 = 100/9.8

Taking the square root of both sides gives:

t = sqrt(100/9.8) ≈ 3.19 seconds

So, it takes approximately 3.19 seconds for the camera to reach the ground.

(b) To find the velocity of the camera just before it lands, we can use the equation:

v = u + gt

where:
v = final velocity
u = initial velocity (0 m/s since the camera is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time (3.19 seconds)

Substituting the values into the equation:

v = 0 + 9.8 * 3.19

v ≈ 31.18 m/s (rounded to two decimal places)

Therefore, the velocity of the camera just before it lands is approximately 31.18 m/s upward.

A hot-air balloon is descending at a rate of 1.5 when a passenger drops a camera.If the camera is 41 above the ground when it is dropped, how long does it take for the camera to reach the ground?

a) The initial velocity is -2.4m/s

hf=ho+Vi*t-4.9t^2
hf=0, hi=50, Vi given, solve for t.
b) Vf=Vi-9.8t