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Projectiles

Projectiles. Courtesy www.physics.ubc.ca/.../p420_00/ darren/web/range/range.html. Two Dimensional Motion. Courtesy Physics Lessons.com. What’s a Projectile?. An object moving in two dimensions under the influence of gravity alone. You Predict.

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Projectiles

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  1. Projectiles Courtesy www.physics.ubc.ca/.../p420_00/ darren/web/range/range.html Two Dimensional Motion Courtesy Physics Lessons.com

  2. What’s a Projectile? • An object moving in two dimensions under the influence of gravity alone.

  3. You Predict • Two identical balls leave the surface of a table at the same time, one essentially dropped the other moving horizontally with a good speed. Which hits the ground first? Courtesy of www.mansfieldct.org/schools/ mms/staff/hand/drop.jpg

  4. Galileo’s Analysis • Horizontal and vertical motions can be analyzed separately • Ball accelerates downward with uniform acceleration –g • Ball moves horizontally with no acceleration • So ball with horizontal velocity reaches ground at same time as one merely dropped.

  5. Courtesy Glenbrook South Physics, Tom Henderson

  6. Horizontal vx = vx0 x = x0 +vx0t Vertical vy = vy0 –gt y = y0 + vy0t – 1/2gt2 vy2 = vy02 -2gy Equations Assuming y positive up; ax = 0, ay = -g = - 9.80 m/s2

  7. Problem Solving • Read carefully • Draw Diagram • Choose xy coordinates and origin • Analyze horizontal and vertical separately • Resolve initial velocity into components • List knowns and unknowns • Remember vx does not change • vy = 0 at top

  8. Examples Horizontal launch • Student runs off 10m high cliff at 5 m/s and lands in water. How far from base of cliff? • y = -1/2gt2 (with y up +) • time to fall t = (2y/-g)1/2 = (-20/-9.80) 1/2 = 1.43 s • x = vx0t = 5 x 1.43 = 7.1 m

  9. Extension • How fast would the student have to run to clear rocks 10m from the base of the same 10m high cliff? • Again t = (2y/-g)1/2 = 1.43 s • x = vx0 t • vx0 = x/t = 10m / 1.43s = 7.0 m/s

  10. Upwardly Launched Projectile With Velocity v and Angle q • vx0 = v0cosq • vy0 = v0sinq • Time in air = 2 vy0/g • Range = R = vx0t = v0cosq x2 vy0/g = (2v02/g) sinqcosq = (v02 /g )sin2q • This is Range Formula • Only for y = y0 v0 q Gunner’s version sin2q = Rg/v02 Trigonometric identity: 2 sinqcosq = sin2q R

  11. Questions • What is the acceleration vector at maximum height? • How would you find the velocity at a given time? • How would you find the height at any time? • How does the speed at launch compare with that just before impact? 9.80 m/s2 downward at all times Use kinematics equations for vx and vy , then find v by sqrt sum of squares of them. Use kinematics equation for y Same

  12. Longest Range • What angle of launch gives the longest range and why? Assume the projectile returns to the height from which launched. • 45 degrees; must maximize sin2q; maximum value of sine is one, happens for 900; then q = 450 • If a launch angle q gives a certain range, what other angle will give the same range and why? • Hint: R goes as sinqcosq

  13. Moving Launch Vehicle • If ball is launched from moving cart, where will it land?

  14. Simulation • Link to simulation • Virtual Lab

  15. A Punt • Football kicked from 1.00 m above ground at 20.0 m/s at angle above horizontal q0 = 370 .Find range. • Can we use range formula? • No. It doesn’t apply since y ≠ y0 • Let x0 = y0 = 0 y = -1.00 m Note: Projectile motion is parabolic Vy0 = v0sinq = 20 x 0.6 = 12m/s

  16. Punt, continued • Find time of flight using y = y0 + vy0t – 1/2gt2 • -1.00m = 0 + (12.0m/s)t – (4.90 m/s2)t2 • (4.90 m/s2)t2 -(12.0m/s)t – (1.00m) = 0 • t = 2.53s or -0.081 s (impossible) • x = vx0t = v0cos370 t= (16.0 m/s)(2.53s) = 40.5 m

  17. Using Formulas • Be sure the formula applies to the situation – that the problem lies within its “range of validity” • Make sure you understand what is going on.

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