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

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Presentation on theme: "Projectiles Two Dimensional Motion Courtesy www.physics.ubc.ca/.../p420_00/ darren/web/range/range.html Courtesy Physics Lessons.com."— Presentation transcript:

1 Projectiles Two Dimensional Motion Courtesy darren/web/range/range.html 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 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 Equations Horizontal v x = v x0 x = x 0 +v x0 t Vertical v y = v y0 –gt y = y 0 + v y0 t – 1/2gt 2 v y 2 = v y0 2 -2gy Assuming y positive up; a x = 0, a y = -g = m/s 2

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 v x does not change v y = 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 = -1/2gt 2 (with y up +) time to fall t = (2y/-g) 1/2 = (-20/-9.80) 1/2 = 1.43 s x = v x0 t = 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 = v x0 t v x0 = x/t = 10m / 1.43s = 7.0 m/s

10 Upwardly Launched Projectile With Velocity v and Angle  v x0 = v 0 cos  v y0 = v 0 sin  Time in air = 2 v y0 /g Range = R = v x0 t = v 0 cos  x  2 v y0 /g = (2v 0 2 /g) sin  cos  v 0 2 /g )sin2  This is Range Formula Only for y = y 0  R Trigonometric identity: 2 sin  cos  sin2  Gunner’s version sin2  Rg/v 0 2 v0v0

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/s 2 downward at all times Use kinematics equations for v x and v y, 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 sin2  maximum value of sine is one, happens for 90 0 ; then  = 45 0 If a launch angle  gives a certain range, what other angle will give the same range and why? Hint: R goes as sin  cos 

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

14 Simulation Link to simulation Link Virtual Lab

15 A Punt Football kicked from 1.00 m above ground at 20.0 m/s at angle above horizontal  0 = Find range. Can we use range formula? No. It doesn’t apply since y ≠ y 0 Let x 0 = y 0 = 0 y = m Note: Projectile motion is parabolic V y0 = v 0 sin  x  12m/s

16 Punt, continued Find time of flight using y = y 0 + v y0 t – 1/2gt m = 0 + (12.0m/s)t – (4.90 m/s 2 )t 2 (4.90 m/s 2 )t 2 - (12.0m/s)t – (1.00m) = 0 t = 2.53s or s (impossible) x = v x0 t = v 0 cos37 0 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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