1. Projectiles calculations Calculating the components of a vector using trigonometry and vertical and horizontal problems

2. Projectiles – vector component calculation Example: If a projectile is launched with an initial velocity of 50m/s at an angle of 60 degrees above the horizontal. What is the horizontal and vertical velocity? V x = V init * cos 60 = 50m/s * cos 60 = 25m/s V y = V init * sin 60 = 50m/s * sin 60 = 43m/s v = 50m/s V y = 43m/s 60° V x = 25m/s

3. Second Sample Problem A water balloon is launched with a speed of 40m/s at an angle of 60° to the horizontal. V x = 40m/s * cos 60° = 20m/s V y = 40m/s * sin 60° = 34.6m/s

4. Time of Flight To determine the time of flight of a projectile use the formula t up = V iy /g Where t up = time to the peak V iy = initial vertical velocity g = gravity (9.8m/s/s) If a projectile has a vertical velocity of 39.2m/s it would take 4 seconds to reach it’s peak t up = 39.2m/s / 9.8m/s/s = 4 sec

5. Equations for horizontal components of motion x = v ix * t + ½ * a x * t 2 V fx = v ix + a x * t V fx 2 = v ix 2 + 2 * a x * x Where x = horizontal displacement m a x = horizontal acceleration m/s/s t = time in s v ix = initial horizontal velocity m/s v fx = final horizontal velocity m/s

6. Equations for vertical components of motion y = v iy * t + ½ *a y * t 2 v fy = v iy + a y * t v fy 2 = v iy 2 + 2 * a y * y Where y = vertical displacement in m a y = vertical acceleration in m/s/s t = time in sec v iy = initial vertical velocity in m/s v fy = final vertical velocity in m/s

7. Sample Problem A pool ball leaves a.60 meter high table with an initial horizontal velocity of 2.4m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table’s edge and ball’s landing location. Known: HorizontalVertical x = ?y = -.60m v ix = 2.4m/sv iy = 0m/s a x = 0m/s/sa y = -9.8m/s/s The pool ball has no horizontal acceleration since it is falling and the y displacement is negative since it is going down and gravity is negative since it is pulling the pool ball down.

8. Sample problem cont. First use the vertical equation y = v it * t + a y * t 2 -.60m = (0m/s) * t +.5 * (-9.8m/s/s) * t 2 -.60m = (-4.9m/s/s) * t 2 0.122s 2 = t 2 t =.350s now it is time to find the horizontal displacement

9. Sample problem cont. x = v ix * t + 0.5 * a x * t 2 x = 2.4m/s *.350s + 0.5 * 0m/s/s*(.350) 2 x =.84m + 0 =.84m So the pool ball is in the air for.35 seconds and lands a horizontal distance of.84m from the pool table