You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?M M FAHAD JOY said:Homework Statement
If a falling object overcome 72 metres in 6 second, how much distance it overcame in first 3 second.
Homework Equations
s = ut+(at^2)/2
The Attempt at a Solution
Here,
s = 72 m
u = 0
t = 6s
a = ?
We know,
s = ut+(at^2)/2
Or, 72 = 0*6 + (a*6^2)/2
Or, 72 = 36a/2
Or, 72 = 18a
Or, a = 4 m/s
In the second part,
u = 0
t = 3s
a = 4 m/s^2
s = ?
Again,
s = ut+(at^2)/2
= 0*t + (4*3^2)/2
= (4*9)/2
= 36/2
= 18 m (ans.)
It's from my school exam question.tnich said:You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?
The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.M M FAHAD JOY said:It's from my school exam question.
It's about falling object on Earth. But the gravity of Earth is 9.8 as I know.
We know that the gravity doesn't depend on mass. But for the friction of air reduces the acceleration of gravity. So it a problem me seeing the first law of falling bodies.
I have checked it again. It is ok.tnich said:The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.
In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?M M FAHAD JOY said:I have checked it again. It is ok.
I am not so well in English. I have translated it from my own language. That's the problem.tnich said:In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?
Do you think the object could have started out going upward, reached a maximum height and then fallen?M M FAHAD JOY said:I am not so well in English. I have translated it from my own language. That's the problem.
I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".tnich said:Do you think the object could have started out going upward, reached a maximum height and then fallen?
Yes, the translation will be fall 72 metres.tnich said:I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".
Then they fell into the trap.M M FAHAD JOY said:Most of my friends answered 36 metres.
Are you sure, quarter of the distance in half time?haruspex said:Then they fell into the trap.
Moving half the distance in half the time would be right for uniform velocity, but for uniform acceleration from rest it is a quadratic, so a quarter of the distance in half the time.
½at2. What happens to that if you halve t? Note, this is only for falling from rest.M M FAHAD JOY said:Are you sure, quarter of the distance in half time?
Ok, Thanks.haruspex said:½at2. What happens to that if you halve t? Note, this is only for falling from rest.