Physics 111. Lecture 8 (Walker: 4.3-5) 2D Motion Examples. Projectile - General Launch Angle. In general, v 0x = v 0 cos θ and v 0y = v 0 sin θ

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1 Physics 111 Lecture 8 (Walker: 4.3-5) D Motion Examples February 13, 009 Lecture 8 1/ Projectile - General Launch Angle In general, v 0x = v 0 cos θ and v 0y = v 0 sin θ (This ASSUMES θ is measured CCW from +x axis!) The equations of motion are: x = v 0x t y = v 0y t - ½gt Lecture 8 /

2 General Launch Angle Trajectory Snapshots of a trajectory; red dots are at t = 1 s, t = s, and t = 3 s Lecture 8 3/ Example: A Home Run Baseball leaves bat making a 30 angle with the ground. It crosses a fence 100 m from home plate at the same height that it was struck. (Neglect air resistance.) What was its velocity as it left the bat? g Lecture 8 4/

3 g v = v cos θ; v = v sin θ; 0x 0 0y 0 x = x + v ( t t ) = ( v cos θ ) t ; 1 0 0x y = 0 = y + v ( t t ) g( t t ) = ( v sin θ ) t gt ; y Solutions: t 1 = 0 or t 1 = v 0 sinθ /g Therefore, x = ( v cos θ) t = ( v cos θ)(v sin θ / g); x = v sinθcos θ / g = v sin θ / g; v = x g = = 0 1 / sin θ (100 m)(9.80 m/s ) / sin(60 ) 33.6 m/s Lecture 8 5/ Trial Question: War at Sea 1 A battleship simultaneously fires two shells at two enemy submarines. The shells are launched with the same initial velocity. If the shells follow the trajectories shown, which submarine gets hit first? (A) Submarine 1 (B) Submarine (C) They are both hit at the same time (D) It depends on the initial velocity Lecture 8 6/

4 Example: To Catch a Thief Police officer chases a jewel thief across rooftops. They are both running when they come to a gap between buildings that is 4.0 m wide and has a drop of 3.0 m. Thief, having studied physics, leaps at 5.0 m/s at an angle of 45 above the horizontal and clears the gap. The police officer did not study physics and thinks he should maximize his horizontal velocity, so he leaps horizontally at 5.0 m/s. (a) Does he clear the gap? (b) By how much does the thief clear the gap? Lecture 8 7/ (a) Does he clear the gap? No! (b) By how much does thief clear the gap? 1 y = y0 + v0yt gt Police: Thief: m = (9.81 m/s )t t = 6.0 m / 9.81 m/s = 0.78 s x= x0 + v0xt = 0 + (5.0 m/s)(0.78 s) = 3.91 m 3.0 m = 0 + (5.0 m/s)sin 45 t (9.81 m/s ) t t = 0.50 s or t = 1. s 1 x= x0 + v0xt = 0 + (5.0 m/s)(1. s)cos 45 = 4.31 m Lecture 8 8/

5 Projectile Motion: Range Range: the horizontal distance a projectile travels If the initial and final elevation are the same: Lecture 8 9/ Projectile Motion: Range Lecture 8 10/

6 Projectile Motion:Maximum Range The range is a maximum when θ = 45 : Lecture 8 11/ Projectile Motion: Symmetry Symmetry in projectile motion: Same y and v Same y and v Lecture 8 1/

7 Path of Projectile y x Lecture 8 13/ ConcepTest A gun is accurately aimed at a dangerous criminal hanging from the gutter of a building. The target is well within the gun s range, but the instant the gun is fired and the bullet moves with a speed v o, the criminal lets go and drops to the ground. What happens? The bullet hits the criminal regardless of the value of v o.. hits the criminal only if v o is large enough. 3. misses the criminal. Lecture 8 14/

8 Example: A Supply Drop Helicopter drops a supply package to flood victims on a raft. When package is released, helicopter is 100 m directly above the raft and flying at velocity 5.0 m/s at angle 36.9 above horizontal. [Neglect air resistance; Sin(36.9 ) = 3 / 5 ; Cos(36.9 ) = 4 / 5 ]. (a) How long is the package in the air? (b) How far from the raft does the package land? (c) If the helicopter continues at constant speed, where is it when the package lands? g Lecture 8 15/ v = v sin θ = (5.0 m/s)(sin36.9 ) = 15.0 m/s; v = v cos θ = 0.0 m/s 0y 0 0 0x yt () = v0 yt gt 1 = gt v t+ y 0 0 y v 0y ± v0y gy (15.0 m/s) ± (15.0 m/s) (9.81 m/s )( 100 m) t = = g (9.81 m/s ) t = 3.4 s and t = 6.30 s x= xh = v0 xt = (0.0 m/s)(6.30 s) = 16 m y = y + v t = 0 + v t = (15.0 m/s)(6.30 s) = 94.4 m h h0 0h 0y g The helicopter will be 194 m above the package impact point. Lecture 8 16/

9 Summary - D Kinematics Components of motion in the x- and y- directions can be treated independently. In projectile motion, the y-component of acceleration is g; there is no x- acceleration. If the launch angle is zero, the initial velocity has only an x-component. The path followed by a projectile is a parabola. The range is the horizontal distance the projectile travels. Lecture 8 17/ Forces Usually think of a force as a push or pull Vector quantity May be contact or field force Lecture 8 18/

10 Fundamental Forces Types Strong Nuclear Force Electromagnetic force Weak nuclear force Gravity Characteristics All field forces Listed in order of decreasing strength Only gravity and electromagnetic in mechanics Lecture 8 19/ Newton s First Law, cont. External force any force that results from the interaction between the object and its environment Statement of Newton s 1st Law: When there is no net external force acting on an object, the acceleration of the object is zero. Lecture 8 0/

11 Inertia and Mass Inertia is the tendency of an object to continue in its original motion Mass is a measure of the inertia, i.e resistance of an object to changes in its motion caused by a force Recall: mass is a scalar quantity SI Unit of mass: kilograms (kg) Lecture 8 1/ End of Lecture 8 Before the next lecture, Walker Homework Assignment #4b should be submitted using WebAssign by 11:00 PM on Monday, February 16. Lecture 8 /