Trajectory - Horizontally Launched Projectiles - pvhelp

The third task of every problem involves the calculation of the x- and y-position values at 1-second intervals. The position at a time of 0.0 seconds is stated in the description to the left of the diagram. Since the projectile is launched horizontally, this given launch velocity is the x-velocity at t = 0.0 seconds. The y-velocity is 0 m/s at t = 0.0 seconds since the projectile has not yet begun to move vertically. Now you will have to determine the x- and y-positions at the other times. Here's how:

Horizontal Position (x)

Projectiles do not accelerate horizontally. They move with a constant horizontal speed. That is they change their position by the same amount every second. If the original horizontal speed (v_ox) is 12 m/s, then every 1 second will result in the projectile changing its x-position by 12 meters. A useful equation that expresses this idea is ...

x = x_o + v_ox • t

where x is the position at time t for an object with an original position of x_o and an original x-velocity of v_ox. The x_o value is 0 m in this problem.

Vertical Position (y)

Projectiles accelerate in the downward direction. The amount of vertical position change is not the same from one second to the next. The easiest way to calculate the vertical position (y) is through the use of the equation ...

y  = y_o + v_oy•t  +  0.5•g•t²

In this problem, the original vertical velocity (v_oy) is 0.0 m/s. and the original y-position (y_o) is 0 m. Thus, 0.5•g•t²  is the expression for calculating the y-position at each of the four times. Since g is a negtive value, the y-positions will all be negative position values. (NOTE: some students might have selected a -10 m/s/s g value).