How does the release of a sandbag affect the motion of a hot air balloon?

[mopar969]

A hot air balloonist rising vertically with a constant velocity of magnitude 5m/s releases a sandbag at an instant when the balloon is 40m above the ground. After it is released the sandbag is in free fall.
a) Compute the position and velocity of the sandbag at 0.25s and 1s after it is released.
b) How many seconds after its release will the bag strike the ground?
c)With what magnitude of velocity does it strike the ground?
d) What is the greatest height above the ground that the sandbag reaches?
e) sketch a-t, v-t, and y-t graphs for the motion.
part a I used x = x +vo(t)+(1/2)a(t^2) and got -0.94 m for 0.25 s and -0.1 m for 1s
also: I used v = vo +at and got -2.55 m/s for 0.25s and 4.8 m/s for 1s
Please check answers and help me get started with what equation to use for part b.

 

[rl.bhat]

A hot air balloonist rising vertically with a constant velocity of magnitude 5m/s releases a sandbag at an instant when the balloon is 40m above the ground. After it is released the sandbag is in free fall.
a) Compute the position and velocity of the sandbag at 0.25s and 1s after it is released.
b) How many seconds after its release will the bag strike the ground?
c)With what magnitude of velocity does it strike the ground?
d) What is the greatest height above the ground that the sandbag reaches?
e) sketch a-t, v-t, and y-t graphs for the motion.
part a I used x = x +vo(t)+(1/2)a(t^2) and got -0.94 m for 0.25 s and -0.1 m for 1s
also: I used v = vo +at and got -2.55 m/s for 0.25s and 4.8 m/s for 1s
Please check answers and help me get started with what equation to use for part b.

Use proper sign convension.
Since the sand bag finally falls on the ground, theequation should be
-x = xo + vo*t - 1/2*g*t^2.

Now proceed.

 

[mopar969]

Okay thank you for the fix my new answers are for 0.25s = 1.56m and for 1s = 9.9m. Is my velocities correct for both times and what equation do I use to find the time in part b.

 

[mopar969]

For the time for the sandbag to hit the ground I got 2.86seconds using(t=sqrt(x/4.9). Also, for part c I got 23.03m/s Is this correct (I used Vf = vi + at). Are these answers correct?

I also just calculated the greatest height above the ground that the sandbag reaches which is 39.75m (I used x = -4.9t^2+2t+x0) Is this correct?

 

[rl.bhat]

For the time for the sandbag to hit the ground, use

-h = vo*t - 1/2*g*t^2. Substitute the values and solve the quadratic to find t.

For c, use -v = vo - gt.

For d, Maximum height H = 40 + v^2/2g

 

[mopar969]

When I solved the quadratic equation where a = -4.9 , b = -5, and c = 0 I got 1.02m? Please help me figure out what I did Wrong. For c I used the equation given and used these values vi= -5 a =9.8 and t = 2.86 and I got an answer of 33.03 m/s. Is this correct?

For d I used the equation given(40+((v^2)/2g)) and used the values v= -5 and g = 9.8 and got an answer of 41.28 m. Is this correct?