A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground. After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
a)Compute the position of the sandbag at a time 0.290 s after its release.
b)Compute the velocity of the sandbag at a time .290s after its release.
e)How many seconds after its release will the bag strike the ground. (i have worked out to be 3.41s.)
f)with what magnitude of velocity does it strike?
G)what is the greatest height above the ground that the sandbag reaches?
i need the answer !!!
To solve this problem, we can use the equations of motion for free fall.
a) Compute the position of the sandbag at a time 0.290 s after its release:
We'll assume that the upward direction is positive, so the initial position of the sandbag is h = 40.0 m. We can use the equation:
y = y0 + v0t + 0.5at^2
where y is the final position, y0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
Substituting the known values, we have:
y = 40.0 m + 0 + 0.5 * -9.8 m/s^2 * (0.290 s)^2
Calculating this expression will give you the position of the sandbag at t = 0.290 s after its release.
b) Compute the velocity of the sandbag at a time 0.290 s after its release:
We can use the equation:
v = v0 + at
where v is the final velocity, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
Substituting the known values, we have:
v = 0 + (-9.8 m/s^2) * 0.290 s
Calculating this expression will give you the velocity of the sandbag at t = 0.290 s after its release.
e) How many seconds after its release will the bag strike the ground:
To find the time it takes for the sandbag to strike the ground, we can use the equation:
y = y0 + v0t + 0.5at^2
Since the final position, y, is 0 (corresponding to the ground), we have:
0 = 40.0 m + 0 + 0.5 * (-9.8 m/s^2) * t^2
This is a quadratic equation in terms of t. Solving this equation will give you the time it takes for the sandbag to hit the ground.
f) With what magnitude of velocity does it strike:
The magnitude of velocity at which the sandbag strikes the ground can be found by calculating the velocity just before it hits the ground. You can use the equation:
v = v0 + at
where v is the final velocity, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
Substituting the known values, we have:
v = 0 + (-9.8 m/s^2) * t
The time, t, can be found in part e). Calculating this expression will give you the magnitude of velocity at which the sandbag strikes the ground.
g) What is the greatest height above the ground that the sandbag reaches:
To find the greatest height above the ground that the sandbag reaches, we can use the equation for vertical displacement:
y = y0 + v0t + 0.5at^2
Since the final velocity is 0 at the highest point, the equation becomes:
0 = 40.0 m + 0 + 0.5 * (-9.8 m/s^2) * t^2
This is a quadratic equation in terms of t. Solving this equation will give you the time it takes for the sandbag to reach its maximum height. Substituting this time back into the equation will give you the greatest height above the ground that the sandbag reaches.
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