- #1
DeanH87
- 2
- 1
- Homework Statement
- Hi, Can someone help with the below calculation. Sample attached
- Relevant Equations
- See Below
Dean
Last edited by a moderator:
What do you know about resolving forces in 2 dimensions?DeanH87 said:Homework Statement:: Hi, Can someone help with the below calculation. Sample attached
Relevant Equations:: See Below
View attachment 264461Top example- How do I get to 31.7 m/s from 30.8 and 7.7? This is way over my head and need help. Thanks in advance
Dean
The formula for calculating the resultant velocity in a triangle is v = √(u2 + w2 - 2uwcosθ), where v is the resultant velocity, u and w are the initial velocities, and θ is the angle between the two initial velocities.
To find the angle between two initial velocities in a triangle, you can use the inverse cosine function: θ = cos-1((u2 + w2 - v2)/2uw). Alternatively, you can use the sine or tangent functions, depending on the given information.
Yes, the resultant velocity in a triangle can be greater than the sum of the initial velocities. This occurs when the angle between the initial velocities is acute, meaning less than 90 degrees. In this case, the resultant velocity will be larger than the sum of the initial velocities.
If the angle between the initial velocities is 180 degrees, then the resultant velocity will be equal to the difference between the two initial velocities. This is because the cosine of 180 degrees is -1, making the formula for resultant velocity v = √(u2 + w2 + 2uw).
Triangle calculation for resultant velocity can be used in various real-life situations, such as calculating the speed and direction of a boat in a river with a current, determining the velocity of a projectile launched at an angle, or finding the velocity of a car moving on a curved road. It is also commonly used in physics and engineering to analyze the motion of objects.