balloon rises w/ v = 16 ft/s, released sandbag at h = 64 ft
- Thread starter bobers
- Start date
Re: integration word problem
Hello, bobers!
\(\displaystyle \text{We're expected to know the free-fall formula: }\;h(t) \;=\;h_o + v_ot - 16t^2\)
. . \(\displaystyle \text{where: }\:\begin{array}{ccc}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ h(t) &=& \text{height at }t\text{ seconds} \end{array}\)
\(\displaystyle \text{We are given: }\:h_o \,=\,64,\;v_o \,=\,16\)
. . \(\displaystyle \text{Hence, the height function is: }\;h(t) \;=\;64 + 16t - 16t^2\)
When does the sandbag strike the ground? . . . when its height is zero.
\(\displaystyle \text{We have: }\:64 + 16t - 16t^2 \:=\:0\)
\(\displaystyle \text{Divide by -16: }\:t^2 - t - 4 \:=\:0\)
\(\displaystyle \text{Quadratic Formula: }\:t \;=\;\frac{1 \pm \sqrt{17}}{2}\)
\(\displaystyle \text{The positive answer is: }\:t \;=\;\frac{1 + \sqrt{17}}{2} \;\approx\;2.56\text{ seconds.}\)
Hello, bobers!
\(\displaystyle \text{We're expected to know the free-fall formula: }\;h(t) \;=\;h_o + v_ot - 16t^2\)
. . \(\displaystyle \text{where: }\:\begin{array}{ccc}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ h(t) &=& \text{height at }t\text{ seconds} \end{array}\)
\(\displaystyle \text{We are given: }\:h_o \,=\,64,\;v_o \,=\,16\)
. . \(\displaystyle \text{Hence, the height function is: }\;h(t) \;=\;64 + 16t - 16t^2\)
When does the sandbag strike the ground? . . . when its height is zero.
\(\displaystyle \text{We have: }\:64 + 16t - 16t^2 \:=\:0\)
\(\displaystyle \text{Divide by -16: }\:t^2 - t - 4 \:=\:0\)
\(\displaystyle \text{Quadratic Formula: }\:t \;=\;\frac{1 \pm \sqrt{17}}{2}\)
\(\displaystyle \text{The positive answer is: }\:t \;=\;\frac{1 + \sqrt{17}}{2} \;\approx\;2.56\text{ seconds.}\)
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Were you told to assume that the balloon rises the same as a rock that is tossed into the air at 16 feet per second?
If not, then I don't know how to determine its acceleration.
Also, balloons released from ground level have an initial velocity of zero.
If the phrase "initial velocity" means the balloon's velocity at ground level, then it must have been released from the bottom of a hole or somehow shot into the air.
It seems to me that the acceleration of this particular rising balloon depends upon the height above sea level from which it's released, the density of the gasses inside the balloon, the mass of the material from which the balloon is made, and the mass of the object attatched the balloon. There may be even more factors of which I'm unaware.
Perhaps, there are a lot of assumptions that go with this exercise, and you did not type them.
If not, then I don't know how to determine its acceleration.
Also, balloons released from ground level have an initial velocity of zero.
If the phrase "initial velocity" means the balloon's velocity at ground level, then it must have been released from the bottom of a hole or somehow shot into the air.
It seems to me that the acceleration of this particular rising balloon depends upon the height above sea level from which it's released, the density of the gasses inside the balloon, the mass of the material from which the balloon is made, and the mass of the object attatched the balloon. There may be even more factors of which I'm unaware.
Perhaps, there are a lot of assumptions that go with this exercise, and you did not type them.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
If the balloon's velocity is constant, then it looks like Soroban did the entire exercise for you.
If the balloon's velocity is constant, then it also means that somebody played a dirty trick on you by using the phrase "initial velocity".
What a strange scenario, overall.
If the balloon's velocity is constant, then it also means that somebody played a dirty trick on you by using the phrase "initial velocity".
What a strange scenario, overall.