A hot air balloon rises at a constant acceleration

[Girn261]

Homework Statement

A hot air balloon is released from the ground & rises up with a constant acceleration of 4 m/s^2. When it is 16 meters above a sandbag is dropped.

What is the time taken for the sandbag to hit the ground?

Homework Equations

v^2=u^2+2as
v=u+at

The Attempt at a Solution

v^2=u^2+2as
v^2=0+2(4)(16)
v=11.31m/s

v=u+at
11.31=0+(4)(t)
t=2.828

 

[BvU]

Would that be the time taken for the sandbag to hit the ground ?
Or is it the time the balloon plus bag take to rise 16 m and reach a speed of 11.3 m/s?

 

[Charles Link]

You found the sandbag is traveling upwards w.r.t. the Earth when it is released at a height of 16 meters. After this, it will accelerate downward from Earth's gravity with $a=-9.8$ m/sec^2. You need to write the equation of its motion, $s$ vs. $t$ starting from the time it is released. The problem has a couple of steps to it, but this $s$ vs. $t$ equation is an important one.

 

[Girn261]

You found the sandbag is traveling upwards w.r.t. the Earth when it is released at a height of 16 meters. After this, it will accelerate downward from Earth's gravity with $a=-9.8$ m/sec^2. You need to write the equation of its motion, $s$ vs. $t$ starting from the time it is released. The problem has a couple of steps to it, but this $s$ vs. $t$ equation is an important one.

s=ut+1/2at^2
16=0+1/2*(9.81)(t)^2
t=1.806s

So I assume this is the time taken to reach the ground after the upward descent from the acceleration of the hot air balloon. But if I add the two together (1.806+2.828) I get the wrong answer.

 

[Charles Link]

s=ut+1/2at^2
16=0+1/2*(9.81)(t)^2
t=1.806s

So I assume this is the time taken to reach the ground after the upward descent from the acceleration of the hot air balloon. But if I add the two together (1.806+2.828) I get the wrong answer.

Your equation in the form you have it is incorrect. The equation has the form $s=s_o+v_o t+(1/2)a t^2$. You need to figure out what $s_o$, $v_o$, and $a$ need to be. Also what is $s$ when the sandbag hits the ground? In addition, the $t$ in this formula is a clock that starts when the sandbag is released. That part I think you already figured out though. $\\$ And additional comment: For the equation of $s$ vs. $t$, there are a couple of options available, e.g. you can choose up as either positive or negative, but you need to be consistent. There is also a choice of where to select the origin for $s$, i.e.the place where $s=0$, but again, you need to be consistent.

 

[Girn261]

Here's my attempt
S=So+(Vo)(t)+1/2(a)(t)^2
0=16+0+(1/2(9.81+4)t^2)
T=-4.79s

I tell you, physics is one of my worst subjects lol. Ugh.. confusing figuring out what's initial and what's final, which acceleration to use. Etc

 

[Charles Link]

Here's my attempt
S=So+(Vo)(t)+1/2(a)(t)^2
0=16+0+(1/2(9.81+4)t^2)
T=-4.79s

I tell you, physics is one of my worst subjects lol. Ugh.. confusing figuring out what's initial and what's final, which acceleration to use. Etc

The sandbag is moving (quickly) upward (w.r.t. the earth) as the rope is cut. Because of this, the $v_o$ is not zero, but rather the number you previously computed. $\\$ Also the acceleration $a$ after the rope is cut is incorrect. It is much simpler than that. The only force on the sandbag after the rope is cut is due to gravity=this should tell you what the acceleration will be.

 

[Girn261]

The sandbag is moving (quickly) upward (w.r.t. the earth) as the rope is cut. Because of this, the $v_o$ is not zero, but rather the number you previously computed. $\\$ Also the acceleration $a$ after the rope is cut is incorrect. It is much simpler than that. The only force on the sandbag after the rope is cut is due to gravity=this should tell you what the acceleration will be.

Thanks for the help boss, I realized my mistake, I plugged the correct units into the formula. And then used the quadratic formula to get my answer.

Cheers