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Projectile Motion - Falling to the Ground
Projectile motion is the path followed by a ball when you throw it and it falls to the ground. You can change how hard you throw it (which changes its beginning speed) and the direction you throw it. Together these give you vi, the initial velocity vector. This vector and the fact that you are on earth uniquely determine the path that the ball will follow.
Vertically, perpendicular to the ground, the ball is accelerated at g (9.8m/s2 if we are at the surface of the earth) towards the ground due to the force of gravity. Horizontally, parallel to the ground, the motion of the ball is not accelerated, but proceeds at a constant speed because once the ball leaves your hand there is no force applied in the horizontal direction. These two motions can be treated independently and the ball's full motion can be recovered by adding the vertical and horizontal motions back together. If you fire a cannonball at an angle θ and a speed v0, you can use two sets of equations, one for horizontal motion and one for vertical motion:
And the cannonball will hit the ground when y = 0 so you could solve the quadratic equation for time. Solving would give two answers and you will need to use logic to figure out which number is the appropriate answer. One answer is that t = 0 (if y0 = 0) which is where the cannonball starts but we are more interested in where it ends up. If we combine the x and y equations to eliminate time, we can see what the path would look like on a graph, taking (x0,y0) = (0,0),
The airplane shown in the figure is to achieve "weightlessness" for a short period of time by flying along a curved path such that it's acceleration components in x direction a x
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