Projectiles calculations

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# Projectiles calculations - PowerPoint PPT Presentation

Projectiles calculations. Calculating the components of a vector using trigonometry and vertical and horizontal problems. Projectiles – vector component calculation.

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### Projectiles calculations

Calculating the components of a vector using trigonometry and vertical and horizontal problems

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?Vx = Vinit * cos 60 = 50m/s * cos 60 = 25m/sVy = Vinit * sin 60 = 50m/s * sin 60 = 43m/s
• v = 50m/s Vy = 43m/s
• 60° Vx = 25m/s
Second Sample Problem
• A water balloon is launched with a speed of 40m/s at an angle of 60° to the horizontal.Vx = 40m/s * cos 60° = 20m/sVy = 40m/s * sin 60° = 34.6m/s
Time of Flight
• To determine the time of flight of a projectile use the formula tup = Viy/g
• Where tup = time to the peak Viy = 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 peaktup = 39.2m/s / 9.8m/s/s = 4 sec
Equations for horizontal components of motion
• x = vix * t + ½ * ax * t2
• Vfx = vix + ax * t
• Vfx2 = vix2 + 2 * ax * x
• Where x = horizontal displacement m ax = horizontal acceleration m/s/s t = time in s vix = initial horizontal velocity m/s vfx = final horizontal velocity m/s
Equations for vertical components of motion
• y = viy * t + ½ *ay * t2vfy = viy + ay * tvfy2 = viy2 + 2 * ay * y
• Where y = vertical displacement in m ay = vertical acceleration in m/s/s t = time in sec viy = initial vertical velocity in m/s vfy = final vertical velocity in m/s
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: Horizontal Verticalx = ? y = -.60m

vix = 2.4m/s viy = 0m/sax = 0m/s/s ay = -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.

Sample problem cont.
• First use the vertical equation
• y = vit * t + ay * t2
• -.60m = (0m/s) * t + .5 * (-9.8m/s/s) * t2
• -.60m = (-4.9m/s/s) * t2
• 0.122s2 = t2
• t = .350s
• now it is time to find the horizontal displacement
Sample problem cont.
• x = vix * t + 0.5 * ax * t2
• 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