![]() Equations are descriptions of physical reality they are not mathematical cranks that you turn so that you put numbers in and get numbers out that correspond to symbols appearing in them. Fg is always -9.8 m/s in a projectile motion problem The y and x components of a projectile motion problem are independent of each other, besides time Initial. Projectile Motion Formula Vx is the velocity (along the x-axis) Vxo is Initial velocity (along the x-axis) Vy is the velocity (along the y-axis) Vyo is. ![]() The problem is with the way you interpreted it. The problem is not that the equation does not work. So when you use this equation you start from $$0=\frac)# must be zero, which is as it should be for the equation to hold and as you have found out. You also know that the projectile needs a certain amount of non-zero time #\Delta t# to come back down to the same height. 4.3 Projectile Motion Projectile motion is the motion of an object subject only to the acceleration of gravity, where the acceleration is constant, as near the surface of Earth. So for the particular problem you treat, you know that #\Delta y = 0# because you have stipulated that the projectile returns to the same height. The kinematic equations for constant acceleration can be written as the vector sum of the constant acceleration equations in the x, y, and z directions. whether the projectile motion takes place on Earth, the Moon, Mars etc. Learn about Projectile Motion here Summary Kinematics is a topic in which we study the motion of a body without worrying about the cause of the motion. air resistance, maximum height (of a projectile). ![]() This is true for all given heights, all initial velocities and all gravitational accelerations, i.e. Predict the trajectory of a horizontally launched projectile using vectors and kinematic equations. Apply kinematic equations and vectors to solve problems involving projectile motion. The equation says that when the projectile is at a given height going up and at time #\Delta t# later returns to the same height going down, the vertical component of the velocity has the same absolute value but different sign. All what I am asking is since all kinematic equations are correct, all should give the same solution. I am not sure if I got your second point.
0 Comments
Leave a Reply. |