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Deled by the parabola y=(-16x²)/(0.434v²)+1.15x+8, where v is the initial velocity of the

User WHol
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Final answer:

By solving for time in horizontal motion and substituting into the vertical motion equation, we prove that a projectile follows a parabolic trajectory described by y=ax+bx², where a and b are constants.

Step-by-step explanation:

The trajectory of a projectile is indeed parabolic, and we can prove this using basic kinematic equations for projectile motion. First, we solve for the time t in the equation x = Voxt, where Vox is the initial velocity component in the x-direction. Then, we substitute t into the equation for the vertical position y = Voyt - (1/2)gt2, where Voy is the initial velocity component in the y-direction and g is the acceleration due to gravity. After substitution, we will arrive at an equation of the form y = ax + bx2, where a and b are constants, describing a parabolic path.

Moreover, knowing the initial vertical velocity Voy = Vo sin θo and the final position y, we can use these values to find the projectile's final velocity and additional characteristics of the parabolic trajectory.

User Aaron Belchamber
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