# Homework Help: Motion in a straight line

1. Sep 14, 2009

### darklich21

A hot air balloon is rising at 16meters/second when it's passenger throws a ball straight up at 22meters/second. How much later does the passenger catch the ball?

From what I understand so far, they want time. From the problem I'm given the initial velocity (Vi) of the ball, and I know my final velocity at some point will be 0m/s because I will catch it. I know I have to use one or several of the kinematic equations, but I'm not quite sure where to start. Can anyone help?

Equations and steps are much appreciated
1. The problem statement, all variables and given/known data

A hot air balloon is rising at 16meters/second when it's passenger throws a ball straight up at 22meters/second. How much later does the passenger catch the ball?

2. Relevant equations
Vf=Vi + at
Df=Di + ViT + 0.5at^2

3. The attempt at a solution

From what I understand so far, they want time. From the problem I'm given the initial velocity (Vi) of the ball, and I know my final velocity at some point will be 0m/s because I will catch it. I know I have to use one or several of the kinematic equations, but I'm not quite sure where to start. Can anyone help?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 14, 2009

### rl.bhat

What is the initial velocity of the thrown upwards?
How much time it takes to reach the maximum height?
At that instant instant what is the distance between ball and balloon?
After that, the balloon rises with uniform velocity, and ball fall freely.

3. Sep 14, 2009

### darklich21

Well, initial velocity of the ball thrown upwards is given in the problem, 22m/s right?

Uh, Disntance, I used Vf^2=Vi^2 + 2ad. I made Vi=22m/s and Vf=0 since the ball is at its peak. I got d=24.69m

the time it takes to reach the maximum height, What equation would I use, is it Y=Y0 + ViT -0.5gt^2?

And where would I go from there?

4. Sep 14, 2009

### rl.bhat

Initially ball is in the balloon, which is rising with a velocity. It is thrown from the rising balloon. Then what is its initial velocity?
You can find the time to reach the maximum height by using
vf = vi - gt.
During this time distance traveled by the balloon is d = vb*t.
Distance traveled by the ball is given by
vf^2 = vi^2 - 2*g*s.
At that instant, the distance between balloon and the ball is (s - d) = y
At hat instant balloon is rising with uniform velocity and ball is falling freely. If they meet after time t, then
distance traveled by the balloon x = vb*t
and distance traveled by the ball is y - x = 0.5* g*t^2
Solve for t.

5. Sep 15, 2009

### darklich21

When talking about the initial velocity of the ball, since it's being throw from the rising balloon, are you saying the initial velocity of the ball is the speed of the rising balloon + the speed at which you throw it?

Last edited: Sep 15, 2009
6. Sep 15, 2009

### rl.bhat

Yes.

7. Sep 16, 2009

### darklich21

Ok apparently I got the answer wrong so let me show you what I did:

initial velocity of the ball when throw from the balloon= 13m/s

time to reach maximum height: Vf= Vi -gt
0=13m/s -9.8t
t= 1.3265s

distance travelled by balloon at this time: d=Vb x t
D= (13m/s)(1.3265s)=17.2449m

distance travelled by ball: Vf^2=Vi^2 +2ad
0=(13m/s)^2 + 2(9.8m/s^2)(d)
d= 8.6224m

difference between balloon and ball is 17.2449m - 8.6224m = 8.6225m

And then, I used the last equation to solve for time: (y-x) = X0 + V0t + 0.5gt^2
8.6225= 0 + 0 + 4.9t^2
t= 1.33s which was apparently wrong.

Any ideas on what I did wrong?

8. Sep 16, 2009

### rl.bhat

initial velocity of the ball when throw from the balloon= 13m/s
How did you get this one?
It should be 16 m/s + 22 m/s

9. Sep 16, 2009

### darklich21

Oops, sorry about that, see the way my homework computer program works is if i get the wrong answer, it changes the values. So if you don't mind, the balloon is rising at 13 m/s and the ball is throw at 21 m/s.

Alright, so I did this:
initial velocity of the ball when thrown from the balloon is 13m/s + 21m/s = 34 m/s

time to reach maximum height: Vf= Vi -gt
0=31m/s -9.8t
t= 3.1633s

distance travelled by balloon at this time: d=Vb x t
D= (13m/s)(3.1633s)=41.1223m

distance travelled by ball: Vf^2=Vi^2 +2ad
0=(34m/s)^2 + 2(9.8m/s^2)(d)
d= 58.9795m

difference between balloon and ball is 58.9795m - 41.1223m = 17.857m

And then, I used the last equation to solve for time: (y-x) = X0 + V0t + 0.5gt^2
17.857m= 0 + 0 + 4.9t^2
t= 1.91s

which was apparently wrong, any ideas?

10. Sep 16, 2009

### rl.bhat

In the problem the velocity of the balloon is given as 16 m/s. You have taken 13 m/s. Which one is correct?

11. Sep 16, 2009

### darklich21

the 13 m/s, sorry about that, the problem switched it's numbers.

And the ball is being thrown up at 21 m/s, not 22m/s. My apologies.

12. Sep 16, 2009

### darklich21

distance traveled by the balloon x = vb*t

I missed that equation, my bad. But isn't that the same equation as d= vb * t that I used earlier in the problem? What's the difference now?

13. Sep 16, 2009

### rl.bhat

When they meet, the displacement of balloon and the ball is the same. Displacement of balloon is d = vb*t
The displacement of the ball is
d = vi*t - 0.5*g*t2
Equate them and solve for t.

14. Sep 16, 2009

### darklich21

Alright so if i equate them to each other, I get the following:

(Vb*t)=ViT -0.5gt^2

Using some algebraic manipulation, i get the following:

T(Vb-Vi)= -4.9T^2

I divide both sides by T and get:

Vb - Vi= -4.9T

Vb= initial velocity of balloon= 13m/s
Vi= initial velocity of ball = 13+21= 34m/s

therefore:

(13-34)=-4.9T
T=4.29seconds

is this right?

15. Sep 16, 2009

### rl.bhat

I think it is right.

16. Sep 16, 2009

### darklich21

ouch, it's not. I really don't know where I went wrong in this problem. Any ideas?

17. Sep 16, 2009

### rl.bhat

No idea. What is the answer.