a person in a hot air balloon ascending vertically at a constant rate of 12m/s drops a rock over the side when the balloon is 40 m above the ground, what is the velocity of the rock just before it hits the ground, and how long will it take the rock to reach the ground

the rock reaches its peak after 12/g s

it ascends at an average rate of
... (12 m/s + 0 m/s) / 2

it then free falls from h

m g h = ½ m v² ... v = √(2 g h)

t = 2 h / v

To find the velocity of the rock just before it hits the ground, we can use the equation of motion:

v² = u² + 2as

Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the rock was initially at rest)
a = acceleration (acceleration due to gravity, approximately 9.8 m/s²)
s = displacement (distance the rock falls, which is the height of the hot air balloon from the ground, 40 m)

Plugging in these values, we get:

v² = 0² + 2 * 9.8 * 40

v² = 2 * 9.8 * 40

v² = 784

Taking the square root of both sides, we find:

v = √784

v = 28 m/s

Therefore, the velocity of the rock just before it hits the ground is 28 m/s.

To find the time taken for the rock to reach the ground, we can use the equation:

s = ut + 0.5at²

Where:
s = displacement (height of the hot air balloon from the ground, 40 m)
u = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, approximately 9.8 m/s²)
t = time taken (unknown)

Plugging in these values, we get:

40 = 0 * t + 0.5 * 9.8 * t²

40 = 4.9t²

Dividing both sides by 4.9, we have:

t² = 8.16

Taking the square root of both sides, we find:

t = √8.16

t ≈ 2.86 seconds

Therefore, it will take approximately 2.86 seconds for the rock to reach the ground.