Engineering Time interval until a ball returns to its orign

[tremain74]

Homework Statement

I have a problem that I am working on from my Engineering Dynamics Book and I want to see if I am on the right path. A person in a balloon rising with a constant velocity of 4 m/s propels a ball upward with velocity of 1.2 m/s relative to the balloon. After what time interval will the ball return to the balloon? Answer: t = 0.245 seconds.

Relevant Equations

I used the equation vy = -9.8*t + v0*sin0(sin of theta).

This is my solution below.

 

[BvU]

Hi,

Do I really see a $\sin(90)=0.89$ in your picture?

(you should typeset the math in your post using $\LaTeX$!)

Then: The balloon goes up with 4 m/s, and the ball is thrown up with 1.2 m/s relative to the balloon.
So how can you write $v_0=4$ m/s ?

What equation are you solving (or rather: not solving, since you get a negative t) ?

$\$

 

[Steve4Physics]

@tremain74, in addition to what @BvU has already said...

$\sin (90 \text { radians}) = 0.894$ approx. You have forgotten to switch your calculator to degrees mode, so it is treating '90' as 90 radians. But in any case, you need to know that $\sin (90^0) = 1$ without using a calculator!

I can’t follow the logic of your working. In addition you have not made clear what your symbols mean, for example:

Is $v_0$:
- the initial velocity of the ball relative to the balloon?
- the initial velocity of the ball relative to the ground?
- the velocity of the balloon?
- something else?

Is $v_y$:
- the velocity of the ball relative to the balloon after some time, $t$?
- the velocity of the ball relative to the ground after some time, $t$?
- the initial velocity of the balloon relative to the ground?
- something else?

If you deliberately change the meaning of a symbol mid-working (you shouldn't!) you need to state this.

Rethink and try again. Here's a big hint:

You are in an elevator rising at a constant velocity. You throw a ball up at a speed of 1.2m/s relative to you and measure the time it takes to come back to you. Can you find the elevator's velocity from your result?

 

[Babadag]

The balloon up speed is 4 m/s constant. Then the space up to
the point of return of ball it is 4*t.
The total space of the ball is (4+1.2)*t-9.8*t^2/2