Time interval until a ball returns to its orign

  • #1
tremain74
12
3
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.

PXL_20241002_163129761.jpg
 
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  • #2
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) ?

##\ ##
 
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  • #3
@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?
 
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FAQ: Time interval until a ball returns to its orign

What factors affect the time interval until a ball returns to its origin?

The time interval until a ball returns to its origin is primarily affected by the initial velocity of the ball, the angle of launch, the height from which it is thrown, and the acceleration due to gravity. Additionally, air resistance can also play a role, especially for lighter balls or those with larger surface areas.

How can I calculate the time interval for a ball thrown vertically?

For a ball thrown vertically upwards, the time interval until it returns to its origin can be calculated using the formula: t = (2 * v₀) / g, where v₀ is the initial velocity and g is the acceleration due to gravity (approximately 9.81 m/s²). The total time is the time taken to reach the highest point multiplied by two, as the ascent and descent durations are equal.

Does the mass of the ball influence the time it takes to return to its origin?

No, the mass of the ball does not influence the time it takes to return to its origin in a vacuum, as all objects fall at the same rate regardless of mass. However, in the presence of air resistance, the shape and surface area may affect the time, but not the mass itself.

What is the effect of launching the ball at an angle?

When a ball is launched at an angle, the time interval until it returns to its origin is influenced by the vertical component of the initial velocity. The time can be calculated by determining the time to reach the peak height (which is influenced by the vertical launch velocity) and then doubling that time for the descent. The launch angle affects the trajectory and maximum height achieved, thus impacting the total time of flight.

How does air resistance impact the time interval for a ball's return?

Air resistance slows down the motion of the ball, which can increase the time interval until it returns to its origin compared to a scenario without air resistance. The effect of air resistance is more pronounced for lighter balls or those with larger surface areas, as they experience more drag, resulting in a longer time to return to the starting point.

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