Projectile motion
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The document discusses projectile motion, defining it as the motion of objects influenced only by gravity. It provides examples, various trajectories, and formulas to calculate the maximum height, total flight time, horizontal range, and final velocity of a projectile launched at specific angles. Additionally, it includes problems to apply the concepts of projectile motion and their corresponding calculations.
![d) Theobject hitstheground at t = t2 = 2 V0 sin(θ) / g (found in part b
above)
Thecomponentsof thevelocity at t aregiven by
Vx = V0 cos(θ) Vy = V0 sin(θ) - g t
Thecomponentsof thevelocity at t = 2 V0 sin(θ) / g aregiven by
Vx = V0 cos(θ) = 20 cos(25°) Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = -
V0 sin(25°)
ThemagitudeV of thevelocity isgiven by
V = √[ Vx
2
+ Vy
2
] = √[ (20 cos(25°))2
+ (- V0sin(25°))2
] = V0 = 20 m/s](https://image.slidesharecdn.com/projectilemotion-170114053400/85/Projectile-motion-16-320.jpg)
Projectile motion
- 1. ProjectileMotion By Prof. Liwayway Memije-Cruz
- 2. Projectile isabody launched at an anglefollowing acurvaturecalled thetrajectory path. an object upon which theonly forceacting is gravity.
- 3. Examplesof Projectiles Thereareavariety of examplesof projectiles. An object dropped from rest isaprojectile(provided that theinfluence of air resistanceisnegligible). An object that isthrown vertically upward isalso aprojectile(provided that the influenceof air resistanceisnegligible). And an object which isthrown upward at an angleto thehorizontal isalso a projectile(provided that theinfluenceof air resistanceis negligible). A projectileisany object that once pro jected or dropped continuesin motion by itsown inertia and is influenced only by thedownward forceof gravity.
- 5. Regardlessof whether aprojectileismoving downwards, upwards, upwardsand rightwards, or downwardsand leftwards, thefree- body diagram of the projectileisstill as depicted in thediagram at theright.
- 6. First caseof trajectoryFirst caseof trajectory Thetrajectory isahalf parabola
- 7. Newton’s First Law of Motion: Newton’s Law ofNewton’s First Law of Motion: Newton’s Law of IneInertiartia
- 9. Second case of trajectorySecond case of trajectory Trajectory isafull parabola
- 10. A projectileisan object upon which theonly forceis gravity. Gravity actsto influencethevertical motion of theprojectile, thuscausing avertical acceleration. Thehorizontal motion of theprojectileistheresult of thetendency of any object in motion to remain in motion at constant velocity. Dueto theabsenceof horizontal forces, aprojectileremainsin motion with aconstant horizontal velocity. Horizontal forces are not required to keep aprojectilemoving horizontally. Theonly forceacting upon aprojectile isgravity!
- 11. Problem 1: An object islaunched at avelocity of 20 m/sin a direction making an angleof 25° upward with the horizontal. a) What isthemaximum height reached by theobject? b) What isthetotal flight time(between launch and touching theground) of theobject? c) What isthehorizontal range(maximum x aboveground) of theobject? d) What isthemagnitudeof thevelocity of theobject just beforeit hitstheground?
- 13. a) What isthemaximum height reached by theobject? Theformulasfor thecomponentsVx and Vy of thevelocity and componentsx and y of thedisplacement aregiven by Vx = V0 cos(θ) Vy = V0 sin(θ) - g t x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2 In theproblem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2 . Theheight of theprojectileisgiven by thecomponent y, and it reachesits maximum valuewhen thecomponent Vy isequal to zero. That iswhen theprojectile changesfrom moving upward to moving downward.(seefigureabove) and also the animation of theprojectile. Vy = V0 sin(θ) - g t = 0 solvefor t t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds Find themaximum height by substituting t by 0.86 secondsin theformulafor y maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters
- 14. b) Thetimeof flight istheinterval of timebetween when projectileislaunched: t1 and when theprojectiletouchesthe ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence V0 sin(θ) t - (1/2) g t2 = 0 Solvefor t t (V0 sin(θ) - (1/2) g t) = 0 two solutions t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g Timeof flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.
- 15. c) In part c) abovewefound thetimeof flight t2 = 2 V0 sin(θ) / g. Thehorizontal rangeisthehorizontal distancegiven by x at t = t2. range= x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ) V0 sin(θ) / g = V0 2 sin(2θ) / g = 202 sin (2(25°)) / 9.8 = 31.26 meters
- 16. d) Theobject hitstheground at t = t2 = 2 V0 sin(θ) / g (found in part b above) Thecomponentsof thevelocity at t aregiven by Vx = V0 cos(θ) Vy = V0 sin(θ) - g t Thecomponentsof thevelocity at t = 2 V0 sin(θ) / g aregiven by Vx = V0 cos(θ) = 20 cos(25°) Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = - V0 sin(25°) ThemagitudeV of thevelocity isgiven by V = √[ Vx 2 + Vy 2 ] = √[ (20 cos(25°))2 + (- V0sin(25°))2 ] = V0 = 20 m/s
- 17. Problem 2: A projectileislaunched from point O at an angleof 22° with an initial velocity of 15 m/sup an inclineplanethat makesan angleof 10° with thehorizontal. Theprojectilehitstheinclineplaneat point M. a) Find thetimeit takesfor theprojectileto hit theinclineplane. b) Find thedistanceOM.