The last of our Range Equation videos. We solve a problem involving two steps and solving for the launch angle, which can be a difficult step for some students. #PhysicsEd #flipclass #flippedclass
Mr. P when I calculated it myself, I recieved 42.045° or approximately 42°, which is a whole degree less than yours. Why is that?? I did it this way: First I solved the Range equation for v^2 and sincebwe know both of tge velocities are equation, I equated the two range equations as below: R 1 / sin( 2 × theta 1) = R 2 / sin( 2 × theta 2) Then I plugged all the values, take the reciprocal and solved for theta 2 and I obtained: Sin( 2 × theta 2) = 0.9946832389 Then theta 2 = 42.045°. Please assist me with this!
Applications of the cross product: 1. finding the area of a triangle, defined by a group of points in 3-D space, where it is a lot more work to find the altitude and base as you usually do for a triangle in 2-D. You can construct vectors along two of the sides, between the corresponding points, with a common point as the origin of both vectors. Then take the cross product. Take the magnitude of the cross product resultant and divide by 2, to get the area of the triangle. 2. Torque. The torque or moment of a force is defined as vector r cross product with the force, where r is a radius vector from the axis of rotation to the point the force is applied. This resolves the line of action of the force, taking in to account its direction and position, to quantify the rotational effort it causes. 3. Angular momentum. Similar concept as torque, except with momentum taking the place of force. 4. Magnetism. The magnetic force on a charged particle of charge q at speed v in magnetic field B, is a cross product of q*v cross B. 5. Testing a vector field to determine if it is conservative or not. There is a calculation in vector calculus called curl, that uses principles from the cross product to calculate. When curl is zero, it means the vector field is conservative. As in, it could be a field that corresponds to a conservative force, such as gravity and electrostatics.
The sin(2*theta) term comes from re-writing sin(theta)*cos(theta) that appears when deriving this equation, such that we only have one instance of theta. We desire to only have one instance of theta, so we can algebraically invert the equation to solve for theta. Unfortunately, inverse sine can only tell us one solution, so we have to be smarter than our calculator to find the other solution. Knowing what we know about the graph of a sine wave, two supplementary angles have the same sine. But for sin(2*theta) this turns out to be two complimentary angles that have the same value of sin(2*theta). Try graphing sin(theta)*cos(theta), and you will notice that it looks just like a sine wave in terms of shape. Albeit, with a squeeze in both directions, such that it is only half as tall (amplitude) and has only half the wavelength. This term ends up being equivalent to sin(2*theta)/2, which appears in the range equation. There is a trig identity of 2*sin(theta)*cos(theta) = sin(2*theta) which enables us to derive this.
Right, I get that, however, I don't remember every having seen a problem which had "projectile motion on an inclined plane". Could you describe what you mean?
this means that imagine you are climbing a hill and then you decided to throw a ball downward or upward and remember that the hill is at an angle with horizontal surface and you are also at an angle with the hill in such situation how would you calculate range on that inclined plane and the angle etc
The last of our Range Equation videos. We solve a problem involving two steps and solving for the launch angle, which can be a difficult step for some students. #PhysicsEd #flipclass #flippedclass
Mr. P when I calculated it myself, I recieved 42.045° or approximately 42°, which is a whole degree less than yours. Why is that?? I did it this way:
First I solved the Range equation for v^2 and sincebwe know both of tge velocities are equation, I equated the two range equations as below:
R 1 / sin( 2 × theta 1) = R 2 / sin( 2 × theta 2)
Then I plugged all the values, take the reciprocal and solved for theta 2 and I obtained:
Sin( 2 × theta 2) = 0.9946832389
Then theta 2 = 42.045°.
Please assist me with this!
sir a random question is coming in my mind and that is what is the practical use of vector cross product?
Applications of the cross product:
1. finding the area of a triangle, defined by a group of points in 3-D space, where it is a lot more work to find the altitude and base as you usually do for a triangle in 2-D. You can construct vectors along two of the sides, between the corresponding points, with a common point as the origin of both vectors. Then take the cross product. Take the magnitude of the cross product resultant and divide by 2, to get the area of the triangle.
2. Torque. The torque or moment of a force is defined as vector r cross product with the force, where r is a radius vector from the axis of rotation to the point the force is applied. This resolves the line of action of the force, taking in to account its direction and position, to quantify the rotational effort it causes.
3. Angular momentum. Similar concept as torque, except with momentum taking the place of force.
4. Magnetism. The magnetic force on a charged particle of charge q at speed v in magnetic field B, is a cross product of q*v cross B.
5. Testing a vector field to determine if it is conservative or not. There is a calculation in vector calculus called curl, that uses principles from the cross product to calculate. When curl is zero, it means the vector field is conservative. As in, it could be a field that corresponds to a conservative force, such as gravity and electrostatics.
Would you please explain more about solving sin(2 Theta)
The sin(2*theta) term comes from re-writing sin(theta)*cos(theta) that appears when deriving this equation, such that we only have one instance of theta. We desire to only have one instance of theta, so we can algebraically invert the equation to solve for theta. Unfortunately, inverse sine can only tell us one solution, so we have to be smarter than our calculator to find the other solution. Knowing what we know about the graph of a sine wave, two supplementary angles have the same sine. But for sin(2*theta) this turns out to be two complimentary angles that have the same value of sin(2*theta).
Try graphing sin(theta)*cos(theta), and you will notice that it looks just like a sine wave in terms of shape. Albeit, with a squeeze in both directions, such that it is only half as tall (amplitude) and has only half the wavelength. This term ends up being equivalent to sin(2*theta)/2, which appears in the range equation. There is a trig identity of 2*sin(theta)*cos(theta) = sin(2*theta) which enables us to derive this.
sir can you please make a video on projectile motion on an inclined plane?
What would the video be about?
do i have to know range for the ap exam?
I do not think memorizing the range equation would be help for any of the AP Physics exams.
@@FlippingPhysics ok thank you
How you make an angle of 43°, I also want to do it,.....
sir i mean that you should make a specific video covering projectile motion on an inclined plane
Right, I get that, however, I don't remember every having seen a problem which had "projectile motion on an inclined plane". Could you describe what you mean?
this means that imagine you are climbing a hill and then you decided to throw a ball downward or upward and remember that the hill is at an angle with horizontal surface and you are also at an angle with the hill in such situation how would you calculate range on that inclined plane and the angle etc
nice sir
thanks