2 Ideal Projectile Motion What do we mean by the word “ideal” here?Assumptions:The projectile behaves like a particle moving in avertical coordinate plane.The only force acting on the projectile during itsflight is the constant force of gravity.The projectile will follow a trajectory that isperfectly parabolic.
3 Ideal Projectile Motion Position, velocity, acceleration due to gravity, and launch angleat t = 0…Initial velocity:Initial position:r = 0 attime t = 0
4 Ideal Projectile Motion Position, velocity, and acceleration at a later time t…Horizontal range
5 Ideal Projectile Motion Newton’s second law of motion: the force acting on theprojectile is equal to the projectile’s mass times its acceleration.For ideal motion, this force is solely the gravitational force:Let’s solve this initial value problem,using initial conditions:and whenFirst integration:Second integration:
7 Ideal Projectile Motion This is the vector equation for ideal projectile motion. Theangle is the projectile’s launch angle (firing angle,angle of elevation), and is the projectile’s initial speed.Break into a pair of scalar equations:These are the parametric equations for ideal projectilemotion. Gravity constants:
8 Ideal Projectile Motion If the ideal projectile is fired from the point insteadof the origin:
9 Practice Problem Ten seconds after firing, the projectile is about A projectile is fired from the origin over horizontal ground atan initial speed of 500 m/sec and a launch angle of 60 degrees.Where will the projectile be 10 sec later?Parametric equations for this situation, evaluated at this time:Ten seconds after firing, the projectile is aboutm in the air and 2500 m downrange
10 Height, Flight Time, and Range The projectile reaches its highest point when its vertical velocitycomponent is zero:Height at this time:
11 Height, Flight Time, and Range To find the total flight time of the projectile, find the time ittakes for the vertical position to equal zero:
12 Height, Flight Time, and Range To find the projectile’s range R, the distance from the origin tothe point of impact on horizontal ground, find the value of xwhen t equals the flight time:Note: The range is largest when or
13 Height, Flight Time, and Range For ideal projectile motion when an object is launched fromthe origin over a horizontal surface with initial speed andlaunch angle :Maximum Height:Flight Time:Range:
14 Practice Problems m sec m Find the maximum height, flight time, and range of a projectilefired from the origin over horizontal ground at an initial speedof 500 m/sec and a launch angle of 60 degrees. Then graph thepath of the projectile.Max Height:mFlight Time:secRange:m
15 Practice Problems Graph window: [0, 25000] by [–2000, 10000] Find the maximum height, flight time, and range of a projectilefired from the origin over horizontal ground at an initial speedof 500 m/sec and a launch angle of 60 degrees. Then graph thepath of the projectile.Parametric equations for the path of the projectile:Graph window: [0, 25000] by [–2000, 10000]
16 Practice ProblemsTo open the 1992 Summer Olympics in Barcelona, bronzemedalist archer Antonio Rebollo lit the Olympic torch with aflaming arrow. Suppose that Rebollo shot the arrow at a heightof 6 ft above ground level 30 yd from the 70-ft-high cauldron,and he wanted the arrow to reach a maximum height exactly4 ft above the center of the cauldron.
17 Practice Problems Express in terms of the initial speed and firing angle .(b) Use ft and the result from part (a) to find thevalue of
18 Practice Problems (c) Find the value of . Values when the arrow reaches maximum height:
19 Practice Problems (d) Find the initial firing angle of the arrow. Put these parts together:
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