A hot air balloonist, rising vertically with a constant speed of 5m/s, releases a sandbag at an instant, the baloon is 40m above the ground.

Question

Compute the position and velocity of the sandbag is at 0.25s and 1s after its realease.
How many seconds after its release will the bag strike the ground?
How fast is it moving as it strikes the ground?
What i sthe greatest height above the ground that the sandbag reaches?

Answer 1

To solve this problem, we need to use the equations of motion for an object in free fall.

First, let's find the position and velocity of the sandbag at 0.25s after its release:
Using the equation for position (s) as a function of time (t) for an object in free fall:
s = ut + (1/2)at^2
where:
s = final position
u = initial velocity (which is 0 in this case since the bag is released)
t = time
a = acceleration due to gravity = 9.8 m/s^2 (assuming we're near the Earth's surface)

At 0.25s after release:
s = 0 + (1/2)(9.8)(0.25)^2
s = 0 + (1/2)(9.8)(0.0625)
s = 0 + 0.30625
s = 0.30625

So, at 0.25s after release, the sandbag is at a position of 0.30625m above the ground.

To find the velocity at 0.25s after release, we can use the equation for velocity (v) as a function of time (t) for an object in free fall:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 in this case since the bag is released)
t = time
a = acceleration due to gravity = 9.8 m/s^2

At 0.25s after release:
v = 0 + (9.8)(0.25)
v = 0 + 2.45
v = 2.45

So, at 0.25s after release, the sandbag has a velocity of 2.45m/s.

Next, let's find the position and velocity of the sandbag at 1s after its release:
At 1s after release:
s = 0 + (1/2)(9.8)(1)^2
s = 0 + (1/2)(9.8)(1)
s = 0 + 4.9
s = 4.9

So, at 1s after release, the sandbag is at a position of 4.9m above the ground.

v = 0 + (9.8)(1)
v = 0 + 9.8
v = 9.8

So, at 1s after release, the sandbag has a velocity of 9.8m/s.

Now, let's find how many seconds after its release the bag will strike the ground:
We know that the initial position (s) when the bag is released is 40m above the ground.

Using the equation for position (s) as a function of time (t) for an object in free fall:
s = ut + (1/2)at^2
where:
s = final position (which is 0 when the bag strikes the ground)
u = initial velocity (which is 0 when the bag is released)
t = time
a = acceleration due to gravity = 9.8 m/s^2

Setting s = 0 and solving for t:
0 = 0 + (1/2)(9.8)t^2
0 = 4.9t^2
t^2 = 0
t = 0

So, the bag will strike the ground at t = 0s after its release.

Now, let's find the velocity of the sandbag as it strikes the ground:
Using the equation for velocity (v) as a function of time (t) for an object in free fall:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 when the bag is released)
t = time
a = acceleration due to gravity = 9.8 m/s^2

At t = 0s:
v = 0 + (9.8)(0)
v = 0

So, the sandbag is not moving (velocity = 0) as it strikes the ground.

Finally, let's find the greatest height above the ground that the sandbag reaches:
The greatest height above the ground occurs when the velocity of the sandbag becomes zero during its upward motion.

Using the equation for velocity (v) as a function of time (t) for an object in free fall:
v = u + at
where:
v = final velocity (which is 0 at the greatest height above the ground)
u = initial velocity (which is 0 when the bag is released)
t = time
a = acceleration due to gravity = 9.8 m/s^2

Setting v = 0 and solving for t:
0 = 0 + (9.8)t
t = 0

So, the greatest height above the ground that the sandbag reaches is 0m, which means it doesn't reach any greater height.