Engineering How High Was the Balloon When the Stone Was Dropped?

[tremain74]

Homework Statement

I have a problem that says a stone is dropped from a balloon that is ascending at a uniform rate of 10 m/s. If it takes the stone 10 s to reach the ground, how high was the balloon at the instant the stone was dropped? The answer is 485m.

Relevant Equations

I am using the projectile motion. ay = -9.8m/s^2 for negative gravity. Vy = -9.8 + V0sin0. V0 is initial velocity. Sin0 is sin of theta. y = -4.9 t^2 + v0*t* Sin0.

ay = -9.8. Vy = -9.8 + v0*Sin0. y = -4.9*t^2 + v0*t*Sin0. By using formula of y, my solution was -4.9(10s)^2 + (10m/s)(10s)Sin0. I assumed that y was equal to 0. Since -4.9 is m/s^2 the answer would be -490m + 100m*Sin0. Therefore I assumeed Sin0 = 490/100.

 

[.Scott]

Your "relevant equation" does not match your problem description.
Your equation is used for a situation where there is a horizontal component to the velocity of the object in free flight.
But the problem you stated does not presume any horizontal motion - nor would it be affected by horizontal motion.
When I solve the problem as stated by you, I do not get 485 meters, nor do I get anything similar to your calculation, nor do I use trigonometry.

 

[hutchphd]

Your "relevant equation" does not match your problem description.

At least not without specification. To use it, you must argue theta to be 90 deg., then it is correct. But it is a good idea to understand and define each element in the equation.....which you do not do. A free body diagram might be a better start .....

 

[.Scott]

I agree (with @hutchphd) . When theta is taken as 90 degrees, then "Sin0" is 1 and the calculation in the second-to-last statement (of @tremain74 ) is correct.
But what is "Therefore I assumed Sin0 = 490/100." supposed to mean? Even if you are trying to force it to the answer provided (the apparently incorrect 485m), that statement would not be true.

 

[Babadag]

Suppose an observer measured the time since the stone began to fall and it was 10 seconds.
The initial velocity of the stone in the opposite direction was 10 m/s and it stopped its ascent in 10/9.8 seconds.
That means the rock flies up 9.8*(10/9.8)^2/2=5.1m
From this point the stone falls in 10 seconds to the ground 9.8*10^2/2=490 m
That means the balloon was at 490-5.1=484.9 m altitude.

 

[WWGD]

Sin0 being $\frac{490}{100} >>1$ , should itself be a warning.

 

[pbuk]

Suppose an observer measured the time since the stone began to fall and it was 10 seconds.
The initial velocity of the stone in the opposite direction was 10 m/s and it stopped its ascent in 10/9.8 seconds.
That means the rock flies up 9.8*(10/9.8)^2/2=5.1m
From this point the stone falls in 10 seconds to the ground 9.8*10^2/2=490 m
That means the balloon was at 490-5.1=484.9 m altitude.

Congratulations on reverse engineering the answer to work out what the question should have been!

The key lies in interpreting the phrase

If it takes the stone 10 s to reach the ground

In order to reach the answer of 485 m you need to interpret this phrase as "if the stone reaches the ground 10 s after it reaches its highest point" or "if the stone reaches the ground 10 s after its speed is zero".

As written by the OP, the answer is 390 m.

But whatever the original question that was set was, the answer clearly has nothing to do with trigonometry - all that is required is the right selection from the SUVAT equations.