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  1. The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion. In this portion of Lesson 2 you will learn how to describe the motion of projectiles numerically.

  2. There is no vertical component in the initial velocity (\(\mathrm{v_0}\)) because the object is launched horizontally. Since the object travels distance \(\mathrm{H}\) in the vertical direction before it hits the ground, we can use the kinematic equation for the vertical motion: \[\mathrm{(y−y_0)=−H=0⋅T−\dfrac{1}{2}gT^2}\]

  3. The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m / s 14.3 m / s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

  4. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain v → v → at final time t , determined in the first part of the example.

  5. Since there is acceleration only in the vertical direction, the velocity in the horizontal direction is constant, being equal to ⁡. The vertical motion of the projectile is the motion of a particle during its free fall.

  6. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v v and θ v θ v at the final time t t determined in the first part of the example.

  7. The vertical velocity changes by -9.8 m/s each second of motion. On the other hand, the horizontal acceleration is 0 m/s/s and the projectile continues with a constant horizontal velocity throughout its entire trajectory.