Physics Help Forum [SOLVED] Confusing Acceleration Questions

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 25th 2009, 07:46 AM #1 Junior Member   Join Date: Oct 2009 Posts: 24 [SOLVED] Confusing Acceleration Questions Q#1 An object is dropped from a height of 10m. It hits the ground in 2 sec. Calculate: a) Average Speed b) Acceleration of the object c) Final Velocity with which it hits the ground ============================== Now I've got all answers correct but I'm a bit confused regarding part b of the above question. For the acceleration of the object shouldn't it be the value of the force of gravity, which in this case is 9.81??? But that's not the correct answer for this as I got the correct answer by using the second equation of motion s = ut + 1/2 at^2 10 = 0 +1/2 4a a = 5 Could you please explain why 9.81 is not the correct answer for this? Because in many similar questions I just used the force of gravity value for acceleration and the answers were correct. I'm confused
 Oct 25th 2009, 07:50 AM #2 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 the problem is for a different planet, whose g is less. On earth , it would take less time to fall.(1.4 seconds to be precise) Last edited by r.samanta; Oct 25th 2009 at 07:54 AM.
Oct 25th 2009, 08:14 AM   #3
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 Originally Posted by unstopabl3 Q#1 An object is dropped from a height of 10m. It hits the ground in 2 sec. Calculate: a) Average Speed b) Acceleration of the object c) Final Velocity with which it hits the ground ============================== Now I've got all answers correct but I'm a bit confused regarding part b of the above question. For the acceleration of the object shouldn't it be the value of the force of gravity, which in this case is 9.81??? But that's not the correct answer for this as I got the correct answer by using the second equation of motion s = ut + 1/2 at^2 10 = 0 +1/2 4a a = 5 Could you please explain why 9.81 is not the correct answer for this? Because in many similar questions I just used the force of gravity value for acceleration and the answers were correct. I'm confused
a) average speed = distance/time= 10/2 = 5m/s
b) acceleration of the object = 10/(t^2) x 2 = 5 m/s^2
c) final velocity = u + at since u is zero use at = 5 x 2 = 10 m/s

9.81 is not the correct answer, they were falling negative but you probably typed in positive on your calculator half way down.

Oct 25th 2009, 08:38 AM   #4
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 Originally Posted by r.samanta the problem is for a different planet, whose g is less. On earth , it would take less time to fall.(1.4 seconds to be precise)
Okay but that's how the question was given to us, how are we supposed to know if the planet is earth or some other planet? I mean, it shouldn't matter what planet you are on, the formulas work the same, unless the value of gravity given is different than 9.81/10 or it's mentioned in the question that the question is related to such and such planet. I mean these kind of questions are quite basic and they don't require us to first calculate if the question or value of gravity is based on earth or some other planet and then solve the rest of the questions accordingly.

For example another question which was given to us is as follows:

Q) A ball is dropped from a height of 20 m. It bounces back with 3/4th speed, with which it hits the ground. Calculate:

a) Final velocity with which it hits the ground first time.

==========================================

In this question I get the correct answer when I take the Acceleration to be 9.81. You see what I mean? Here I didn't even have to calculate the value of Acceleration first because I thought it's a freely falling object so it's acceleration must be 9.81. And I was correct, it gives the correct answer.

Now this question is quite similar to the first question, then why did we have to calculate a value for acceleration in the first one, which is obviously a different value than the value in this question.

I'm so confused!

Oct 25th 2009, 08:42 AM   #5
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 Originally Posted by Paul46 a) average speed = distance/time= 10/2 = 5m/s b) acceleration of the object = 10/(t^2) x 2 = 5 m/s^2 c) final velocity = u + at since u is zero use at = 5 x 2 = 10 m/s 9.81 is not the correct answer, they were falling negative but you probably typed in positive on your calculator half way down.
Like I said earlier, I've got the correct answers for all of the above. What I didn't understand was why did we have to use a formula to calculate the value of acceleration, when we know that all freely falling object fall with an acceleration of 9.81, I know only on earth, but that's not mentioned in the question whether this is earth or some other planet, so naturally we assume it is earth. I've used 9.81 for a lot of such questions as an acceleration value and got the correct answers. So does that mean this question is flawed?

Secondly how exactly did you solve the part (b) of my first question? I'm not sure which formula you have used. Please elaborate.

Thanks!

 Oct 25th 2009, 08:56 AM #6 Banned   Join Date: Aug 2009 Location: UK Posts: 240 I wouldn't get too worried about getting confused, physics is very complicated indeed, i'm 46 with a good maths background and took interest in physics a few months ago, i'm like you i get confused too. The best thing if your confused is relax and tackle something your confused about slowly and practice makes perfect. with regard the questions you say are flawed, don't think like that, try to break them down with the information they provide, not the intuition you know. with regard my answer displacement (s) = ut + (1/2at^2) since u is zero then i could omit ut so 10 = (1/2)at^2. I used 10/(t^2) = (1/2)a so 10/(t^2) x2 = a i hope you stick around the forum.
 Oct 25th 2009, 09:11 AM #7 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 sorry for late reply. you dont have to worry what the value of g is. if nothing is said, you will naturally take the experiment to be done on earth and use g=9.8. in the problem u posted, the body fell in 2 seconds a distance of 10 metres which is never possible in earth unless there are other resistive forces in operation to reduce the effective g.(it takes 1.4 seconds if u put g=9.8) since in the problem, u are given distance and time taken, u can easily find acceleration whether it is on earth or not. so why worry about acc here?in other problems where u require g and nothing is specified, u take the experiment to be done on earth and use g=9.8.physics is always an exact science.
 Oct 25th 2009, 10:36 AM #8 Junior Member   Join Date: Oct 2009 Posts: 24 Thanks guys, you guys have enlightened me I'm definitely going to stick around These are great forums with great helpers! Keep up the good work.

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