Solving Motion & Speed Homework Problem with X, u and a

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A stone is dropped from a balloon ascending at 5 m/s from a height of 50 m, and the problem requires calculating the stone's height upon impact with the ground and the time taken for the fall. The key equations involve using kinematic formulas, particularly s = ut + 0.5at², while considering the balloon's changing height during the stone's descent. It is emphasized that the stone's fall will take longer than if it were simply dropped due to the balloon's upward motion. The discussion suggests solving the problem both sequentially and simultaneously to gain a comprehensive understanding. Small insights into the physics of the situation can aid in answering related objective questions effectively.
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Homework Statement


A stone is droped from a balloon going up with a uniform velocity of 5m/s.if the baloon was at 50m high when the stone was dropped, find its height when the stone hits the ground?
find the time for

X(distance)=-50m, u(initial velocity)=5m/s a(acceleration)=-10m/s*s

Homework Equations



please show the the calculaion and methods to solve this problem
rather than this method
also if there is any other simpler way to solve this question?

The Attempt at a Solution


i am not able to solve by using X=ut+1/2at*t
i tried the method shown in the book but i m not able to solve the above equation
 
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That's the correct method
Make sure you got the signs correct for each quantity, pick a direction (ie up is +ve)

As a starting point, imagine if the rock was simply dropped - work out the time for that, then you know your answer should be a bit larger than this,
 
Take care of two things:

1. You have to find the height of the balloon.
2. Is the height of the balloon constant? NO! It changes with time. So, you have to find out the height of the balloon at some specific time! What is that time?

Now, I ask another question. What question needs to be solved first? 1 or 2?

For 2, all you have to do is use the s = ut + 0.5at^2

u = -5m/s, a = 10m/s^2, s = height of the balloon when the stone was thrown (+50 or -50?)

Find time.

Then for 1, there's no acceleration! So which equation will you use?
 
Another way to solve the question would be to setup equations for both the cases, with s1 and t1 as the unkowns.

Then solve them simultaneously through the methods learned in ninth standard!

Solve the question both ways in your notebook - simultaneously and sequentially!
 
Think about mgb_physics' suggestions.

He's giving you a physical picture. I gave you a way to get the answer - that's a technique. What he said is, our case will take longer time than the case where the rock would have been simply dropped.

Why's that so? Think about it, draw a simple diagram in which you contrast the two situations. Then tell me what you think!

IMP : SUCH SMALL INSIGHTS CAN BE VERY HELPFUL IN ELIMINATING OPTIONS IN OBJECTIVE QUESTIONS!
 
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