on this problem any hints (formulas) to help me solve this problem would be appreciated. Thanks in advance
am I wrong?TheSilence said:So I need to find the time and range of the projectile.
[tex]V_0x[/tex]= 45m/s cos(45)
[tex]V_0y[/tex]=45m/s sin(45)
and Range is: [tex]R = 2{V_0x}{V_oy}/{g}[/tex]
Time is: [tex]t = {2V_0y}/{g}[/tex]
!Two dimensional motion refers to the movement of an object in two perpendicular directions, typically represented by the x and y axes. It takes into account both the magnitude and direction of an object's displacement.
The key principles of two dimensional motion include displacement, velocity, acceleration, and projectile motion. These principles describe how an object moves in two dimensions and can be used to predict its future motion.
Distance refers to the total length of the path an object has traveled, while displacement refers to the straight line distance between an object's starting and ending points. In two dimensional motion, displacement takes into account both the horizontal and vertical components of an object's movement, while distance does not.
Velocity in two dimensional motion is calculated by dividing an object's displacement in each direction by the time it took to travel that distance. This results in a vector quantity with both magnitude and direction.
Projectile motion is a type of two dimensional motion in which an object is launched into the air at an angle and follows a curved path, influenced by both its initial velocity and the force of gravity. This type of motion is commonly seen in sports like basketball and in everyday activities like throwing a ball.