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.
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?
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.
To solve these problems, we can use the equations of motion, specifically the equations of uniform acceleration.
First, let's find the position and velocity of the sandbag at 0.25s and 1s after its release.
1. Position at 0.25s:
To find the position, we use the equation:
Position = Initial Position + Initial Velocity * Time + (1/2) * Acceleration * Time^2.
Since the balloonist is rising vertically, the initial position is 40m, the initial velocity is 0 (as the sandbag is released), and there is no acceleration (since the speed is constant).
Position at 0.25s = 40m + 0 * 0.25s + (1/2) * 0 * (0.25s)^2 = 40m
Therefore, the position of the sandbag at 0.25s is 40m above the ground.
2. Velocity at 0.25s:
To find the velocity, we use the equation:
Velocity = Initial Velocity + Acceleration * Time.
Again, since the sandbag is released, the initial velocity is 0, and there is no acceleration.
Velocity at 0.25s = 0 + 0 * 0.25s = 0
Therefore, the velocity of the sandbag at 0.25s is 0 m/s (it momentarily stops).
3. Position at 1s:
Using the same position equation:
Position at 1s = 40m + 0 * 1s + (1/2) * 0 * (1s)^2 = 40m
The position of the sandbag at 1s is also 40m above the ground.
4. Velocity at 1s:
Using the same velocity equation:
Velocity at 1s = 0 + 0 * 1s = 0
The velocity of the sandbag at 1s is also 0 m/s (it remains stationary).
Now, let's find how many seconds after its release the bag strikes the ground.
To find the time taken for the sandbag to reach the ground, we can use the second equation of motion:
Position = Initial Position + Initial Velocity * Time + (1/2) * Acceleration * Time^2.
In this case, the initial position is 40m, the initial velocity is 0, and the acceleration is due to gravity, which is approximately -9.8 m/s^2 (negative because it acts downwards).
We want to find the time when the position of the sandbag is 0 (ground level). So the equation becomes:
0 = 40m + 0 * Time + (1/2) * (-9.8 m/s^2) * Time^2
Simplifying the equation:
4.9 * Time^2 = 40
Solving for Time:
Time^2 = 40 / 4.9
Time = √(8.16)
Time ≈ 2.86 seconds
Therefore, the sandbag will strike the ground approximately 2.86 seconds after its release.
To find the speed at which the sandbag strikes the ground, we can use the first equation of motion:
Velocity = Initial Velocity + Acceleration * Time.
The initial velocity of the sandbag is 0, and the acceleration is due to gravity (-9.8 m/s^2).
Velocity = 0 + (-9.8 m/s^2) * 2.86s = -28.05 m/s (negative because it's moving downward)
Therefore, the sandbag is moving at approximately 28.05 m/s as it strikes the ground.
Lastly, let's find the greatest height above the ground that the sandbag reaches.
To find the maximum height, we need to find the time when the velocity of the sandbag becomes 0. At this point, it will momentarily stop before reversing direction.
Using the same equation as before for velocity:
Velocity = Initial Velocity + Acceleration * Time = 0
Solving for Time:
0 = 0 + (-9.8 m/s^2) * Time
Time = 0
Therefore, the sandbag reaches its highest point (greatest height) at Time = 0 (initially).
Hence, the greatest height above the ground that the sandbag reaches is 40m (initial height when released).
To compute the position and velocity of the sandbag at different time intervals after its release, we can use the equations of motion.
Given:
Initial velocity of the sandbag = 0 m/s (as it was released from a stationary position)
Acceleration due to gravity = 9.8 m/s^2 (acting downwards)
1) To find the position and velocity at 0.25 seconds after release:
Time, t = 0.25 s
Using the equation for position:
Position, s = initial position + (initial velocity * time) + (0.5 * acceleration * time^2)
s = 40 m + (0 m/s * 0.25 s) + (0.5 * 9.8 m/s^2 * 0.25 s^2)
s = 40 m + 0 m + 0.5 m
s = 40.5 m
The sandbag is at a position of 40.5 meters above the ground at 0.25 seconds after release.
Using the equation for velocity:
Velocity, v = initial velocity + (acceleration * time)
v = 0 m/s + (9.8 m/s^2 * 0.25 s)
v = 2.45 m/s
The sandbag has a velocity of 2.45 m/s at 0.25 seconds after release.
2) To find the position and velocity at 1 second after release:
Time, t = 1 s
Using the equation for position:
s = 40 m + (0 m/s * 1 s) + (0.5 * 9.8 m/s^2 * 1 s^2)
s = 40 m + 0 m + 4.9 m
s = 44.9 m
The sandbag is at a position of 44.9 meters above the ground at 1 second after release.
Using the equation for velocity:
v = 0 m/s + (9.8 m/s^2 * 1 s)
v = 9.8 m/s
The sandbag has a velocity of 9.8 m/s at 1 second after release.
3) To find the time it takes for the sandbag to reach the ground:
The equation for position can be rearranged to solve for time as follows:
s = 40 m + (0 m/s * t) + (0.5 * 9.8 m/s^2 * t^2)
0.5 * 9.8 m/s^2 * t^2 = -40 m
4.9 m/s^2 * t^2 = -40 m
Ignoring the negative root, we get:
t^2 = -40 m / 4.9 m/s^2
t^2 = 8.1633 s^2
Taking the square root:
t = √(8.1633 s^2)
t ≈ 2.86 s
Therefore, the sandbag will strike the ground approximately 2.86 seconds after its release.
4) To find the speed at which the sandbag strikes the ground:
Using the equation for velocity:
v = 0 m/s + (9.8 m/s^2 * 2.86 s)
v ≈ 28 m/s
The sandbag is moving at approximately 28 m/s as it strikes the ground.
5) To find the greatest height above the ground that the sandbag reaches:
At the highest point, the velocity of the sandbag will be 0 m/s.
Using the equation for velocity:
v = 0 m/s + (9.8 m/s^2 * t)
0 = 9.8 m/s^2 * t
Therefore, the sandbag will reach its greatest height above the ground at t = 0 seconds (immediately after release). The initial height of the sandbag was 40 meters, so the greatest height reached by the sandbag above the ground is 40 meters.