A body initially moving with a speed of 20m/s covers a distance of 150m after 5 sec .What is the acceleration of the body?

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

sergeinik [125]

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 ) ) ]

3 0

3 years ago

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A worker uses a cart to move a load of bricks weighing 680 N a distance of 10 m across a parking lot.

NNADVOKAT [17]

When you are finding work, the easiest way is to use the formula.

W = F*D

Where F is the force and D is the distance. Simply take the constant force of 209N and multiply it by the distance of 10m. Which will give you 2090J

4 0

3 years ago

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A ball rolls down a hill and hits a box. The momentum of the ball decreases. Which happens to the momentum? Select one of the op

Irina-Kira [14]

So we want to know what happens to the momentum of the ball that rolls down hill and hits a box. So we need to use the law of conservation of momentum which states that the momentum must be conserved. It cant be transformed into inertia or mass. It can only be transferred to other object via some interactions like collisions. So it has to be a. transferred to the box and that is the correct answer.

7 0

3 years ago

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$