Final answer:
The trajectory of a projectile is shown to be parabolic by solving for time t using the equation for horizontal motion x = Voxt, substituting it into the vertical motion equation, and simplifying to y = ax + bx², which is a quadratic equation in terms of x.
Step-by-step explanation:
To show that the trajectory of a projectile is parabolic and takes the form y = ax + bx², we start with the standard equations of motion for a projectile moving under the influence of gravity in a vacuum. We assume no air resistance for simplicity.
The horizontal position x of a projectile is given by the equation x = V0xt, where V0x is the initial horizontal velocity and t is time. For the vertical position y, we use the equation y = V0yt - 1/2 g t², where V0y is the initial vertical velocity and g is the acceleration due to gravity. To combine these into a single equation for y in terms of x, we solve for t in the first equation and substitute it into the second.
Solving for t gives us t = x / V0x. Substituting into the y equation, we obtain y = (V0y / V0x)x - (1/2 g / V0x²)x². This simplifies to y = ax + bx² where a = V0y / V0x and b = -1/2 g / V0x², which is the quadratic equation for a parabola concerning x. This parabolic trajectory shows that the motion of a projectile under these conditions will follow a path determined by its initial velocities, both horizontally and vertically, conforming to the described quadratic relationship.