Unit Kinematics Module Investigating Motion in Two Dimensions page 1 of 2 Understanding Projectile Motion Projectile motion in the x direction is independent of projectile motion in the ydirection The time required for a projectile to land at the same height at which it began is v iy g 2 v2 sin 2 The maximal g range occurs when the object is projected at an angle of 45 The range of an object projected from the ground is When you are working problems with projectile motion you can consider the x components independently from the y components The graph on the left shows the path of a projectile The horizontal component of the velocity vector is constant and the x component of the acceleration is zero The vertical component of the velocity increases with time The y component of the m acceleration is 9 8 2 downwards The object s s speed is the length of its velocity vector speed v x 2 v y 2 The speed of a projectile increases with time An object is launched from the ground at an angle with initial speed v You want to calculate the horizontal distance the object travels or its range R The equations of projectile motion that allow you to calculate xf require that you know the travel time so you must first find t You know all the variables in the second equation of projectile motion in the y direction except t Choose a coordinate system so that y 0 is ground level and the positive y direction is upward The object starts and ends at ground level y i y f 0 The initial velocity in the y direction is v sin and the acceleration in the y direction is the acceleration m due to gravity or g 9 8 2 The time required s v iy for the projectile to land is g 2 www thinkwell com info thinkwell com Copyright 2011 Thinkwell Corp All Rights Reserved 1819 doc rev 06 23 2011 Unit Kinematics Module Investigating Motion in Two Dimensions page 2 of 2 Understanding Projectile Motion After you have found the time required for the projectile to land you can use the second equation of projectile motion in the x direction to solve for the range If you choose a coordinate system so that the object is launched from x 0 then xf R The projectile is launched at an angle with initial speed v The object s horizontal component of velocity is v cos and its vertical component is v sin The trigonometric identity 2cos sin sin2 allows you to simplify the equation The range of a projectile that lands at the same height from which it was v2 launched is R sin 2 g If the projectile is launched at an angle of 0 it will be driven directly into the ground and the range is zero If the projectile is launched vertically it will land in the same location from which it began and the range is zero The maximum range occurs when the projectile is launched at an angle of 45 In this case the range v2 is 1 g www thinkwell com info thinkwell com Copyright 2011 Thinkwell Corp All Rights Reserved 1819 doc rev 06 23 2011
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