A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00m/s , releases a sandbag at an instant when the balloon is a height v = 40.0m 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.
g= 9.8 m/s^2
A)Compute the position of the sandbag at a time 0.170 s after its release.
B)Compute the velocity of the sandbag at a time 0.170 s after its release.
C)Compute the position of the sandbag at a time 1.00s after its release.
D)Compute the velocity of the sandbag at a time 1.00 s after its release.
E)How many seconds after its release will the bag strike the ground?
F)With what magnitude of velocity does it strike?
G)What is the greatest height above the ground that the sandbag reaches?
Okay, I have the acceleratoin, the velocity and the distance. What equation do I use?
g= 9.8 m/s^2
A)Compute the position of the sandbag at a time 0.170 s after its release.
B)Compute the velocity of the sandbag at a time 0.170 s after its release.
C)Compute the position of the sandbag at a time 1.00s after its release.
D)Compute the velocity of the sandbag at a time 1.00 s after its release.
E)How many seconds after its release will the bag strike the ground?
F)With what magnitude of velocity does it strike?
G)What is the greatest height above the ground that the sandbag reaches?
Okay, I have the acceleratoin, the velocity and the distance. What equation do I use?
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Use the equations
y = yi + vi*t - ½g*t²
v = vi - g*t
y = yi + vi*t - ½g*t²
v = vi - g*t
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Hi pene, Well, first of all, if upward displacement is positive, then g = -9.8 m/s^2, not +9.8 m/s^2.
The equations to use are:
v^2 = u^2 + 2 x a x s
s = ut + 0.5 x a x t^2
v = u + a x t
Where u = initial velocity (+5 m/s)
v = velocity at time t
s = displacement at time t
t = time (seconds)
NOTE: s is the displacement from the initial position (t = 0) when the sandbag is released, not the displacement from the ground. To get the displacement from the ground, you will need to add 40m
The equations to use are:
v^2 = u^2 + 2 x a x s
s = ut + 0.5 x a x t^2
v = u + a x t
Where u = initial velocity (+5 m/s)
v = velocity at time t
s = displacement at time t
t = time (seconds)
NOTE: s is the displacement from the initial position (t = 0) when the sandbag is released, not the displacement from the ground. To get the displacement from the ground, you will need to add 40m
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Do your own homework.