Kinematics and projection

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Homework Statement


Hi
I have been stumped by this question for the past few days. Worrying since I will be sitting STEP in June! Never mind. It goes like this:
A particle is projected vertically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.
These are my parameters:
From A to B
u=30
a=-9.8
s=h
From B
u=?
a=-9.8
s=?
t=2.4
I took t=2.4 as the Maximum height above B.(I am not sure about this line of reasoning!)
Based on that V=0
Using V=U + at
U=23.52 m/s (This seemed to make sense since particle will start decelerating at some point)
So, U (at b)=23.52 m/s
Considering it from A to B
U=30
V=23.52
a=-9.8
S=h
Using V^2=U^s +2aS
h=17.7m (3s.f)
But alas, it seems my answer (and probable reasoning) is wrong. Any suggestions???


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
cepheid
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I took t=2.4 as the Maximum height above B.(I am not sure about this line of reasoning!)
I assume that you meant to say, "I took t = 2.4 s to be the elapsed time at which the particle reached its maximum height above point B." Otherwise, the statement is just nonsense.

In any case, this line of reasoning is wrong. 2.4 s is not the amount of time required for the particle to go up from point B to its maximum height. To see why not, re-read the problem statement carefully. It states that the particle spends 2.4 s above point B. That suggests that, 2.4 s after going above point B, the particle has ceased to be above it. So it can't be at the max height after 2.4 s, because then it would STILL be above B.

Q. What must be true if the particle was above B but is no longer above it 2.4 s later (i.e. where is the particle after those 2.4 s)?

A. _________________________________

The answer to this question tells you what point in the particle has reached in its motion at this instant.

Based on that V=0
Using V=U + at
U=23.52 m/s
EDIT: I messed up my critique of this part of your solution (so I deleted it and replaced it with this sentence), but the basic problem is that v is not 0 at t = 2.4 s after reaching B, so what you have calculated is not the speed at point B.

(This seemed to make sense since particle will start decelerating at some point)
The particle is accelerating downward continuously throughout the motion! That is why the speed steadily decreases on the way up, which is why it eventually stops and comes back down.

The above were my comments on your solution. Now here is my advice on how to go about solving it:

Basically, you can divide the motion into two phases. (1) Everything that happened after reaching point B, and (2) everything that happened before reaching point B. You can use the given info to tell you everything you need to know about (1), which in turn allows you to solve for (2) completely.

In particular, since you know how much time the particle spends above B, you can figure out what its velocity must have been at point B. From that point on, you're off to the races, because knowing the difference in velocities between points A and B allows you to determine what the distance between them must have been.

EDIT 2: I see that this is basically what you did in the first place, only with the wrong v. So, see the beginning of my post for an explanation of why your v was wrong.
 
Last edited:
  • #3
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I assume that you meant to say, "I took t = 2.4 s to be the elapsed time at which the particle reached its maximum height above point B." Otherwise, the statement is just nonsense.

In any case, this line of reasoning is wrong. 2.4 s is not the amount of time required for the particle to go up from point B to its maximum height. To see why not, re-read the problem statement carefully. It states that the particle spends 2.4 s above point B. That suggests that, 2.4 s after going above point B, the particle has ceased to be above it. So it can't be at the max height after 2.4 s, because then it would STILL be above B.

Q. What must be true if the particle was above B but is no longer above it 2.4 s later (i.e. where is the particle after those 2.4 s)?

A. _________________________________

The answer to this question tells you what point in the particle has reached in its motion at this instant.



EDIT: I messed up my critique of this part of your solution (so I deleted it and replaced it with this sentence), but the basic problem is that v is not 0 at t = 2.4 s after reaching B, so what you have calculated is not the speed at point B.



The particle is accelerating downward continuously throughout the motion! That is why the speed steadily decreases on the way up, which is why it eventually stops and comes back down.

The above were my comments on your solution. Now here is my advice on how to go about solving it:

Basically, you can divide the motion into two phases. (1) Everything that happened after reaching point B, and (2) everything that happened before reaching point B. You can use the given info to tell you everything you need to know about (1), which in turn allows you to solve for (2) completely.

In particular, since you know how much time the particle spends above B, you can figure out what its velocity must have been at point B. From that point on, you're off to the races, because knowing the difference in velocities between points A and B allows you to determine what the distance between them must have been.

EDIT 2: I see that this is basically what you did in the first place, only with the wrong v. So, see the beginning of my post for an explanation of why your v was wrong.
Thanks. But very strong critique though!o:) Please make your tone softer next time. I work full time and study independently. So, its all in my head and my reasoning. Me, myself and I.

But I appreciate it. As you can guess, I am typing from work now. Will give you a feedback after work.
 

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