Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Physics problem on freely falling bodies.

Here is the problem:

"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  View Figure  . 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."

Question:

Compute the position of the sandbag at a time 0.295 s after its release.

We are supposed to be doing this in some online math software and you input the answer and it validates it for you etc. I solved it but when I put my answer it says its incorrect... Here what i did:

y = y0 + V0yt + .5Ayt^2

y = final position (find)
y0 = initial position (40)
V0y = initial velocity (5 m/s)
t = time (0.295)
Ay = acceleration (gravity) = -9.8 m/s

So given the information in problem, I have to give the final position of the sandbox which corresponds to y.

y = 40 + (5 m/s)(0.295) + .5(-9.8 m/s)(0.295)^2
y = 40 +1.475 - 0.5(9.8)(0.87025)
y = 41.475 - 4.264225
y = 37.210775
y ~ 37.21

However, the online thing keeps saying its wrong? Could anyone help me? I have to turn this in by 12.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Physics problem on freely falling bodies.

I believe the third line of your working should read: y = 41.475 - 0.4264225.
It's also possible that the online thing wants your answer to 3 significant figures, given that all the numbers in the question are that.

---

Why did the vector cross the road?
It wanted to be normal.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Physics problem on freely falling bodies.

0.5(9.8)(0.87025) = 4.264225 ?

Nevermind. Your right!

Thanks.

Last edited by LuisRodg (2008-01-13 06:45:23)