Antiderivatives Blind!

[Asimov Pupil]

I'm afraid i am not seeing something help would be apreciated.

 

you toss a stone upwards off a 63m cliff at 8 meters per second

how long til it hits the ground?

 

here is what i have

 

[math]a=-9.8\frac{m}{s}[/math]

 

then the velocity using antiderivatives is

 

[math]v=-9.8t=8[/math]

 

and distance

 

[math]s=-4.9t^2+8t+63[/math]

 

then if [math]s=0[/math]

 

[math]t=4.49sec[/math]

 

my question is that if you throw a stone upwards off a cliff then wouldn't the C value in the s formula be higher than that?

 

what am i missing?

[Johnny5]

I'm afraid i am not seeing something help would be apreciated.

 

you toss a stone upwards off a 63m cliff at 8 meters per second

how long til it hits the ground?

 

Unfortunately' date=' latex isn't working.

 

Let's see...

 

The simplest way to do this, is to treat the event set theoretically.

 

i will have to explain, because I am one of few who do this.

 

Here is your given information:

 

Cliff height above ground=63 meters

Initial speed= 8 meters per second.

Initial direction of projectile motion=straight up

 

Start the clock at the moment you release the projectile.

 

There is a moment in time at which the object comes to rest in the frame.

 

That is one subevent, and lasts for time T1.

 

There is a second subevent, which begins when the first subevent ends, and ends when the object strikes the ground.

 

Denote the amount of time of the second subevent by T2.

 

The whole is the union of the parts. So the time of the whole event is T1+T2.

 

First compute T1, then compute T2, then add them to obtain the answer.

 

Here is the kinematical formula for constant acceleration:

 

D= v[sub']0[/sub]t + 1/2 a t²

 

The acceleration due to gravity has magnitude 9.8 meters per second squared, and its direction is opposite to the direction of motion.

 

So for subevent one we have:

 

D= 8t - (9.8/2) t²

 

Now, as you can see this is one equation in one unknown.

 

The height that the object rises wasn't given, nor was the time of flight t.

 

So we need a second equation in the same two unknowns, and then we can solve for T1, which is the amount of time between when the stone is released, to the moment in time at which it comes to rest in the frame.

 

Here is a link to the kinematical equations for constant acceleration:

 

Look and you will see this one;

 

D = (v_i+v_f)t/2

 

We know the initial speed is 8 m/s

And we know the final speed is 0 m/s

 

Keep in mind that speed is a strictly non-negative quantity (distance traveled divided by time of travel).

 

Ok so...

 

Using the formula above we have:

 

D = 8t/2=4t

 

And we also have:

 

D= 8t - (9.8/2) t²

 

 

So we have two equations in two unknowns, so we can now solve for both unknowns. It is the time of the first subevent that we are concerned with, so we need to eliminate D.

 

By direct substitution we have:

 

4t= 8t - (9.8/2) t²

 

9.8/2 = 4.9 therefore

 

4t= 8t - 4.9t²

 

Hence:

4.9t²+ 4t-8t=0

 

Hence:

 

4.9t²+ -4t=0

 

Factor out a t to obtain:

 

t(4.9t -4) =0

 

One root is zero, which cannot be an amount of time, because an amount of time is a strictly positive quantity, hence we want the other root, namely the one for which:

 

4.9t -4 =0

 

So:

 

4.9t = 4

 

So:

 

t = 4/4.9 = .81 seconds

 

So the amount of time of the first subevent is .81 seconds.

 

T1=.81 seconds

 

Now, we have to compute the amount of time for the second subevent.

 

During this part of the whole event, the object falls to the earth from higher than 63 meters.

 

Let D be the height the object rose above the cliff during the first subevent.

 

Then the distance which the object falls during the second part, is 63+D.

 

We need to know that in order to compute the time of the second subevent T2.

 

We already know this:

 

D =4t

 

So T1=.81 hence

 

D = 4(.81) = 3.26 meters

 

H = 63 + 3.26 = 66.26

 

The amount of time of the second subevent satisfies the following equation:

 

H = v₀t + (9.8/2) t²

 

Now during this part of the whole event, the initial speed is zero. The magnitude of the acceleration hasn't changed, but the direction of motion has. So during the first subevent, the center of mass of the object was decelerating in the frame, during the second subevent, the center of mass is accelerating in the frame.

 

So the initial speed is zero, hence:

 

H = (9.8/2) t²

 

Hence:

 

66.26 = (9.8/2) t²

 

Hence:

 

66.26 = 4.9 t²

 

Hence:

 

13.52 = t²

 

Hence:

 

3.67 = t

 

So the time of the second subevent is 3.67 seconds:

 

T2 = 3.67 seconds

 

And

 

T1 = .81 seconds

 

So

 

T=T1+T2 = 3.67+.81 = 4.48 seconds

 

Unless I made an error. You can compare with other peoples calculations.

 

Regards

 

 

PS: The thing about breaking the whole event into mutually exclusive and collectively exhaustive parts, is because:

 

1. It leads to the right answer.

2. It will help you analyze problems in special relativity theory.

[Aeternus]

I know this is the analysis/calculus forum but looking at it from a purely mechanics point of view its pretty nice. One equation, resulting in a quadratic which you can solve for the answer (as D or S is displacement rather than distance and so can take a minus value).

 

So in mechanics terms -

 

( Taking Up at Positive Direction)

S= -63 ( Displacement )

U=+8 ( Initial Velocity )

A=-9.8 ( Acceleration )

T=? ( Time )

 

s = ut + 1/2at^2

 

so -63 = 8t + (-9.8/2)t^2

 

-4.8t^2 + 8t +63 = 0

 

( -8 +/- sqrt(64 - (4x-4.9x63) ) / -9.8

 

( -8 +/- 36.04 ) / -9.8 so we need a positive time so we need a negative on the top to cancel the negative on the bottom.

 

-8 - 36.04 = -44.04 -> -44.04/-9.8 = 4.49 s = 4.5s

 

Although, im guessing what your talking about it more complicated than that (sounds like it) and has to be worked out differently and I also could have made a mistake, just felt like working through it.