Projectiles calculations
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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

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Projectiles calculations

Projectiles calculations

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


Projectiles vector component calculation

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

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

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

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

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

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: HorizontalVerticalx = ?y = -.60m

    vix = 2.4m/sviy = 0m/sax = 0m/s/say = -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

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 cont1

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


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