Kinematics in one dimension - hot air balloon and bullet

[slc23]

Homework Statement

A hot air balloon is ascending straight up at a constant speed of 8.10 m/s. When the balloon is 13.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 28.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places? Enter the answers in the ascending order.

Homework Equations

Kinematic equations:
V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

The Attempt at a Solution

I tried using the X - xo = Vot + .5at^2 equation for both the balloon and the bullet to get two variables, delta x and t, and then solving for each. But I've tried it several times and I keep getting funky answers (like negative values for t) and I don't know what I'm doing wrong.

 

[cepheid]

Welcome to PF,

Can you post what you tried? What was your equation for the height vs time of the balloon? And of the bullet?

What do you have to do with these equations in order to figure out what the problem is asking for?

 

[slc23]

Sure :)

So in my latest attempt --

Balloon
V = 8.10 m/s
a = 0 m/s^2

x = Vot + .5at^2
x=(8.10)t

(I am assuming there is no acceleration because velocity is constant, and I am putting in the velocity of the balloon as initial velocity, not sure if this is a correct assumption or not...)

Bullet
Vo = 28 m/s
a = -9.8 m/s^2 (acceleration due to gravity, in this problem I am assuming up is positive so acceleration is negative)

x = Vot + .5at^2
x=(28)t - 4.9t^2

Setting them equal to each other...

8.1t = 28t - 4.9t^2
4.9t^2 - 19.9t = 0
t (4.9 t - 19.9) = 0 giving one solution t=0
4.9t - 19.9 = 0
4.9 t = 19.9
t = 4.06 s

Then I subbed into the balloon equation..

x = 8.1 t
x = (8.1)(4.06)
x = 32.9
and x = 0 when t=0, but adding the 13 m of the balloon height giving
x = 13, x = 32.9

So I know it's wrong and I probably made a bunch of different mistakes, but I am feeling reallyyyy lost

Thanks!

 

[cepheid]

Sure :)

So in my latest attempt --

Balloon
V = 8.10 m/s
a = 0 m/s^2

x = Vot + .5at^2
x=(8.10)t

(I am assuming there is no acceleration because velocity is constant, and I am putting in the velocity of the balloon as initial velocity, not sure if this is a correct assumption or not...)

Yes, all of your assumptions are, of course, correct. However, it would probably be easiest if you chose t = 0 to be the moment at which the bullet is fired. Therefore, the "initial" position of the balloon is not 0 m, but rather 13.0 m. In other words, we have a non-zero x₀ term in the equation:

x = x₀ + v₀t

where x₀ = 13.0 m.

Bullet
Vo = 28 m/s
a = -9.8 m/s^2 (acceleration due to gravity, in this problem I am assuming up is positive so acceleration is negative)

x = Vot + .5at^2
x=(28)t - 4.9t^2

This looks okay to me as well. Try solving it again, this time taking into account the initial height of the balloon, which you forgot.

 

[slc23]

Worked perfectly -thank you very much!