Projectile Motion, General Solution What follows is a general solution for the two dimensional motion of an object thrown in a gravitational field. This is usually termed a projectile motion problem. The thrown object is called the projectile.

Its path is called the trajectory. We will answer all the usual questions that arise in a first year physics class regarding this motion. We will not consider air resistance.

Autocad p and id examples. Without air resistance, the projectile will follow a parabolic trajectory. We will be throwing the projectile on level ground on planet Earth. It will leave the point of release, arc through the air along a path shaped like a parabola, and then hit ground a certain distance from where it was thrown.

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As mentioned above, this is a two dimensional problem. Therefore, we will consider x and y directed displacements, velocities, and accelerations. The projectile will accelerate under the influence of gravity, so its y acceleration will be downward, or negative, and will be equal in size to the acceleration due to gravity on Earth. There will be no acceleration in the x direction since the force of gravity does not act along this axis.

On Earth the acceleration due to gravity is 9.8 m/s 2 directed downward. So, for this presentation acceleration in the y direction, or a y, will be -9.8 m/s 2, and acceleration in the x direction, or a x, will be 0.0 m/s 2. Given the original conditions with which the projectile is thrown we will proceed to find the components of the original velocity and then move on to answer the following questions: • How much time passes till the projectile is at the top of its flight? • How high does the projectile rise? • How much time passes till the projectile strikes the ground? • How far away does the projectile land from its starting point?

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Original, or initial, conditions: The original conditions are the size of the velocity and the angle above the horizontal with which the projectile is thrown. General: Original size of velocity: v o Original angle: theta Example: v o = 40.0 m/s theta = 35 degrees Components of original velocity: The usual first step in this investigation is to find the x and y components for the original velocity. General: X component of original velocity: v ox = v ocos(theta) Y component of original velocity: v oy = v osin(theta) Example: In the x direction: v ox = v ocos(theta) v ox = (40.0 m/s)(cos(35 degrees)) v ox = (40.0)(0.8191) v ox = 32.76 v ox = 32.8 m/s In the y direction: v oy = v osin(theta) v oy = (40.0 m/s)(sin(35 degrees)) v oy = (40.0)(0.5735) v oy = 22.94 v oy = 22.9 m/s How much time passes until the projectile is at the top of its trajectory?