Physics -- Part 2

I have another quick question. It looks easy, but for some reason I just don't think I'm doing it right.

Please check over this:

"A kid throws a ball straight up into the air. The ball is caught 2.5 seconds later. "

a) How high did the ball go (from its point of release)?
b) What was the speed of the ball the moment before it was caught?





The first thing that threw me off was no initial velocity was given... I assume it's 9.8 m/s?

My solutions:

a) x = .5(vi+vf)(t)
x = .5(9.8)(2.5/2) = 6.125 m

I divided the time since you can conside both the up and down direction symmetrical.

I believe that is correct, but not sure about this b part:

b) vf = vi + at
vf = 9.8(2.5/2) = 12.25

What do you do here when there is no initial velocity? Something just doesn't make since about this problem.





The next question is this:

"A rock is thrown downward from a bridge with a speed of 15 m/s. If the bridge is 50m above a river, how long will it take for the rock to hit the water?"

My solution:

x = vi(t) + .5a(t)^2

-50 = 15t + .5(-9.8)t^2

Solve as quadratic (after multiplying by negative one):

4.9t^2-15t-50 = 0

formula:

t = 15 +- sqrt(15^2 - 4(4.9)(-50) )
--------------------------------------
9.8

and you get: 5.07 seconds... Is that correct?




Any help/clarity would be much appreciated.


PS: These aren't homework questions. My teacher doesn't give homework and that means I have to do some serious studying to prepare for tests.

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Originally Posted by System_Error

I have another quick question. It looks easy, but for some reason I just don't think I'm doing it right.

Please check over this:

"A kid throws a ball straight up into the air. The ball is caught 2.5 seconds later. "

a) How high did the ball go (from its point of release)?
b) What was the speed of the ball the moment before it was caught?





The first thing that threw me off was no initial velocity was given... I assume it's 9.8 m/s?

My solutions:

a) x = .5(vi+vf)(t)
x = .5(9.8)(2.5/2) = 6.125 m

I divided the time since you can conside both the up and down direction symmetrical.

I believe that is correct, but not sure about this b part:

b) vf = vi + at
vf = 9.8(2.5/2) = 12.25

What do you do here when there is no initial velocity? Something just doesn't make since about this problem.

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I think your first part is ok. That the time taken to go up and comming back to the same plase are same.

I'll try to check the second part.

I think the second part should be like this. Please check it correct or not.

I applied the equation to downwards,

so,

S = Ut + 0.5*f*(t^2)

Since I put the equation to downwards,

S = 50m (not -50m isn't it), U = 15m/s and f = 9.8m/s^2 ( not -9.8 isn't it)

After sustitution you get the equation,

9.8*t^2 + 30*t - 100 = 0

Solve this,

I got 2.01S,

The initial velocity can be calculated from its time in the air...

If the ball was travelling at 9.8 ms when it was thrown then it would have been airborn for 2 seconds not 2.5. (ball goes up for 1 second and then falls for 1 second)

Thus the initial (vertical) velocity was (2.5/2) * 9.8 = 12.25m/s

(Dividing by 2 because its a round trip, throw up and fall down. the ball us under constant accelaration by gravity.)

ps. assuming we are ignoring air resistance, the speed of the ball when it is caught is exactly the same as when it was thrown. since it takes gravity 1.25 seconds to bring the ball to a halt at the top of the trajectory, and then another 1.25 seconds for it to accelerate down back through the same distance before it hits the hand again.

Quote:

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Originally Posted by System_Error

The first thing that threw me off was no initial velocity was given... I assume it's 9.8 m/s?

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You cannot assume this...

You can use the equation vi = vf + at to find the initial and final speed (which will be equal to each other). Since direction matters, vi = -vf, so:

vi = -vi + at,

vi + vi = at

2vi = (9.8 x 2.5)

vi = 12.25.

You now have vi (and vf) so you can use x = vit + 0.5at² to get the distance the ball travelled after 1.25 seconds. I get 7.66m.

Thanks guys. I haven't had time to ask my teacher if he could look over anything, but I believe Dross' approach for the first problem is correct for part b.


As for my second problem... I was completely wrong. This morning I worked it a different way and got the same answer as eranga.


Thanks guys. The help is greatly appreciated!

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Originally Posted by wossname

ps. assuming we are ignoring air resistance, the speed of the ball when it is caught is exactly the same as when it was thrown.

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That's right. If T = 2.5 s, then you can calculate the initial velocity from the fact that when the ball reaches its highest point its velocity is zero:

0 = v = v₀ + at = v₀ - g(T/2)

v₀ = g(T/2) = 9.8 m/s² x (2.5 s/2) = 12.25 m/s (this is the intial velocity and that just before the ball is caught, the answer to part b)

As for the maximum height,

h = v₀t +(1/2)at² = v₀(T/2) -(1/2)g(T/2)²

Rather than substitute the actual value for v₀ we can use g(T/2):

h = g(T/2)(T/2) -(1/2)g(T/2)² = (1/2)g(T/2)² = (1/2) 9.8 m/s² (2.5 s / 2) = 7.66 m

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Originally Posted by eranga262154

I got 2.01S,

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:thumb:

Incidentally, an important point to remember is that you can always resolve these equations into any given direction. As long as you makes sure that all the values you use are with reference to that particular direction (either positive or negative), then they will work. These equations are all basically equivalent - it just depends on the information you are given.

So, for example, you were wondering about the initial speed of the ball.

Using s=ut+0.5at^2 and resolving in the vertical direction, taking upwards to be the positive direction:

u = (s - 0.5at^2)/t ...(rearranging the above)

s = 0 m (because the ball ends up back where it started, in the boy's hands)
a = -9.8 ms^-2 (because upwards is positive here)
t = 2.5 s

u = 0.5 * 9.8 * 2.5 = 12.25 ms^-1.

This technique is important to remember, and will enable you to solve pretty much any such question.

Another typical example is the cannonball question, fired at some angle. You can still resolve this vertically and horizontally - remembering that vertically s=0 and horizontally a=0 - to get all the rest of the parameters.



zaza

Yeh... there are five useful formulas for projectile physics (Zaza stated 1) using the folowing:
t = time (sec)
s = distance (m)
u = initial velocity (m/s)
v = final velocity (m/s)
a = acceleration (m/s/s)

1. s/t = (u+v)/2
2. v = u+at
3. 2as = v^2-u^2
4. s = vt - (1/2)at^2
5. s = ut + (1/2)at^2

These can be used and rearranged to find any of the values, there are just a few things to remember; you can always asume a = 9.8 m/s/s (depends on the question), ignoring wind resistance, the motion is usually symetrical (again depends on the equation) and therefore it is usually easier to split the motion in two, velocity and acceleration are vector quantities, meaning that they have a direction. That means if you throw something up, and gravity will be going in the opposite direction which also means that either velocity or acceleration will have a positive value, and the other will have a negative value (in your case the velocity would be positive and the acceleration would be negative).

Thats about all I can think of.

When you get horizontal movement as well as vertical movement it gets a bit more interesting. Just remember to split the equation into vertical and horizontal. also calculating the velocity from an angle (the sum of the vector velocity horizontal and vector velocity vetical). But I wont get into that.

To solve moving (point) object problems, the best thing to do is just learn Newton's second law, F = ma where F is the force, m the mass and a the acceleration (second time derivative of the position). Then it all amounts to know the force as a function of the position (which is often given), the velocity or whatever, and solve (integrate) the resulting differential equation whereby all formulae are derived.

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Originally Posted by krtxmrtz

To solve moving (point) object problems, the best thing to do is just learn Newton's second law, F = ma where F is the force, m the mass and a the acceleration (second time derivative of the position). Then it all amounts to know the force as a function of the position (which is often given), the velocity or whatever, and solve (integrate) the resulting differential equation whereby all formulae are derived.

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Doing it that way requires an understanding of momentum, which with some rearrangement and a few other formula, leads to the five formula I mentioned. I personally dislike using derivatives for projectile motion. It seems too messy, and there is no real understanding there.

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Originally Posted by Prodian

Doing it that way requires an understanding of momentum, which with some rearrangement and a few other formula, leads to the five formula I mentioned. I personally dislike using derivatives for projectile motion. It seems too messy, and there is no real understanding there.

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The point is, knowing Newton's law and calculus (the latter is a must for physics) you don't need to memorize any formulas -though after some use you end up by learning them anyway.
And calculus isn't messy, on the contrary, it does make sense and it's one of the greatest achievements of mankind. It's THE calculation tool by excellence. It's like having a Swiss army knife. :)

Quote:

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Originally Posted by krtxmrtz

The point is, knowing Newton's law and calculus (the latter is a must for physics) you don't need to memorize any formulas

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This statement is profound and right on. Last year I did some tutoring of intro physics students and was amazed at the obsession with trying to pick which forumula to 'plug the numbers' into and the resulting confusion - no understanding of the fundamentals at all. To me, nothing is simpler than drawing the free body diagram, writing the basic equations of motion, integrating twice, and solving for what's needed.

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Originally Posted by VBAhack

...the obsession with trying to pick which forumula to 'plug the numbers' into and the resulting confusion - no understanding of the fundamentals at all.

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This type of mentality along with its causes and consequences is precisely the main topic this book is about. Its a lot of fun to read, even by those people easily scared by the bare mention of the word "math".

Quote:

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Originally Posted by VBAhack

This statement is profound and right on. Last year I did some tutoring of intro physics students and was amazed at the obsession with trying to pick which forumula to 'plug the numbers' into and the resulting confusion - no understanding of the fundamentals at all. To me, nothing is simpler than drawing the free body diagram, writing the basic equations of motion, integrating twice, and solving for what's needed.

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To really understand you need to where the formula's came from. They give you a proper understanding of how it is measured and how it works together. For example, the SI units, Distance (m) Time (sec) give you Velocity (m/s), which gives you how many metres travelled for everyone second. Aceleration gives you how many metres per second that the velocity increases every second (m/s/s).

When you use calculus, you don't know physics (the real part behind maths), you know calculus.

Quote:

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Originally Posted by Prodian

To really understand you need to where the formula's came from. They give you a proper understanding of how it is measured and how it works together. For example, the SI units, Distance (m) Time (sec) give you Velocity (m/s), which gives you how many metres travelled for everyone second. Aceleration gives you how many metres per second that the velocity increases every second (m/s/s).

When you use calculus, you don't know physics (the real part behind maths), you know calculus.

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You're quite right, to know and undertand physics it's not enough to know calculus. However, calculus is so profoundly related to the way physics systems work that you can almost derive some of the physics from it. For example, you can arrive to the concept of (instantaneous) velocity from the knowledge of average velocity, some vector analysis and the concept of derivative -especially its graphical interpretation.

Of course, there is probably no "absolute truth" about all this and the answer, as usual, lies somewhere in between.

Quote:

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Originally Posted by krtxmrtz

You're quite right, to know and undertand physics it's not enough to know calculus. However, calculus is so profoundly related to the way physics systems work that you can almost derive some of the physics from it. For example, you can arrive to the concept of (instantaneous) velocity from the knowledge of average velocity, some vector analysis and the concept of derivative -especially its graphical interpretation.

Of course, there is probably no "absolute truth" about all this and the answer, as usual, lies somewhere in between.

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This is a nice definition for Calculus and Physics. I strongly agreed with you.

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Originally Posted by eranga262154

This is a nice definition for Calculus and Physics. I strongly agreed with you.

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Long live calculus, long live Newton, Bernouilli, Euler, Gauss, Cauchy, Riemann...! :thumb:

or really,

those people should Long Live

Here are a few calculus quotes by famous people:

I particularly like these 2:

"For since the fabric of the universe is most perfect and the work of a most
wise Creator, nothing at all takes place in the universe in which some rule
of maximum or minimum does not appear."
L. Euler

"Everyone who understands the subject will agree that even the basis on
which the scientific explanation of nature rests, is intelligible only to those
who have learned at least the elements of the differential and integral calculus,
as well as of analytical geometry."
F. Klein