This is a question on my science hw.

Physics

2 answers:

TEA [102]2 years ago

6 0

I think it is better to prove it,

when you put two things in the same altitude , only when they have the same speed, can fall on the grownd in the same time,

so, we can prove it :

1)we know that the acceleration of the earf (gravity or g)is the same for every thing on it :

we know: V=at +V1 and here it is

V=gt+V1

so Va=Vb because gta +V1a=gtb + V1b

and V1 is 0 =>gta=gtb =>ta=tb

(a and b are names of two assumed things)

pogonyaev2 years ago

5 0

gravity is like a gas that sticks us to ground no matter what object is falling they were both fall down at the same time because it doesn't matter the mass it just matters like the curve around them like if the curve and like their sides

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I have read that it is enough energy to power the needs of civilization for an entire year.

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A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The

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Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be:
L = R * m * vi * cos(90 - theta) 

cos(90 - theta) is just sin(theta) 
and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g 

so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g 

We can combine the two vi terms and get: 

L = vi^3 * m * sin(theta) * sin(2*theta) / g 

What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. 

Now, for the first part... 

There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. 

Okay, so let's get back to what we know: 

L = d * m * v * cos(phi) 

where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... 

m is still m. 
v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) 
d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g 

That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. 

The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: 

h = 1/2 * (vi * sin(theta))^2 / g 

So, there's the rise. So, our *slope* is rise/run, so 

slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] 

The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: 

slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] 

(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). 

Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... 

slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 

Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... 

alpha = arctan( tan(theta) / 4 ) 

Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha 

so, phi = 180 - arctan( tan(theta) / 4 ) 

Now, we go back to our original formula and plug it ALL in... 

L = d * m * v * cos(phi) 

becomes... 

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] 

Now, cos(180 - something) = cos(something), so we can simplify a little bit... 

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ]

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