PHY183-Lecture07 - Physics for Scientists &...

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 7 Maximum Height and Range January 26, 2011 Physics for Scientists&Engineers 1 1 1 Projectile Motion Analysis In the movie below, we toss a ball and observe its trajectory with respect to an x-y coordinate system in cm y (cm) x (cm) January 26, 2011 Physics for Scientists&Engineers 1 2 2 Projectile Motion Analysis (2) We extract the position of the ball for each time frame and plot the results Frame 1 t=0s Frame 7 t = 0.2 s Frame 14 t = 0.47 s Frame 21 t = 0.67 s January 26, 2011 Physics for Scientists&Engineers 1 3 3 Clicker Quiz For the motion of the ball in the x direction we observe that A) the component of the velocity in the x-direction is 264 cm/s B) the component of the velocity in in the x-direction is zero C) the component of the velocity in the x-direction is increasing D) the component of the velocity in the x-direction is 75.1 cm/s January 26, 2011 Physics for Scientists&Engineers 1 4 4 Clicker Quiz Answer For the motion of the ball in the x direction we observe that D) The component of the velocity in the x-direction is 75.1 cm/s January 26, 2011 Physics for Scientists&Engineers 1 5 5 Clicker Quiz For the motion of the ball in the y direction we observe that A) the acceleration in the ydirection is 516 cm/s2 B) the acceleration in the ydirection is -516 cm/s2 C) the acceleration in the ydirection is 1032 cm/s2 D) The acceleration in the ydirection is -1032 cm/s2 January 26, 2011 Physics for Scientists&Engineers 1 6 6 Clicker Quiz Answer For the motion of the ball in the y direction we observe that D) The acceleration in the ydirection is -1032 cm/s2 12 y = y0 + vy 0t − gt 2 Should have gotten -981 cm/s2 January 26, 2011 Physics for Scientists&Engineers 1 7 7 Projectile Motion Analysis (2) We can calculate the velocity as a function of time January 26, 2011 Physics for Scientists&Engineers 1 8 8 Maximum Height of Trajectory Yesterday we derived y = y0 + vy 0 vx 0 x− g2 x 2 2 vx 0 Introduce 2 2 v0 = vx 0 + vy 0 tan θ 0 = vy 0 vx 0 cos θ 0 = vx 0 v0 Result: y = y0 + x tan θ 0 − g x2 2 2 v0 cos2 θ 0 Physics for Scientists&Engineers 1 9 January 26, 2011 9 Maximum Height of Trajectory Starting point: trajectory g y = y0 + x tan θ 0 − 2 x2 2 v0 cos2 θ 0 Take derivative of y with respect to x: dy d g g = y0 + x tan θ 0 − 2 x 2 = tan θ 0 − 2 x 2 2 dx dx 2 v0 cos θ 0 v0 cos θ 0 Find root of this derivative to determine extremum: g 0 = tan θ 0 − 2 xH 2 v0 cos θ 0 2 2 2 v0 cos 2 θ 0 tan θ 0 v0 v0 xH = = sin θ 0 cosθ 0 = sin 2θ 0 g g 2g January 26, 2011 Physics for Scientists&Engineers 1 10 10 Maximum Height of Trajectory (2) Result: 2 v0 xH = sin 2θ 0 2g Convince ourselves that 2nd derivative is smaller than 0: d2y d g g = tan θ 0 − 2 x =− 2 <0 2 2 2 dx dx v0 cos θ 0 v0 cos θ 0 xH is indeed the point where maximum height is reached January 26, 2011 Physics for Scientists&Engineers 1 11 11 Maximum Height of Trajectory (3) What is the maximum height? Answer: H = y ( xH ) = y0 + xH tan θ 0 − = y0 + 2 0 g 2 xH 2 2 2v0 cos θ 0 2 0 2 v v g sin 2θ 0 tan θ 0 − 2 sin 2θ 0 2 2g 2v0 cos θ 0 2 g 2 2 2 v0 v0 v0 2 2 = y0 + sin θ 0 − sin θ 0 = y0 + sin 2 θ 0 g 2g 2g Since vy 0 = v0 sin θ 0 , we finally find: H = y0 + 2 vy 0 2g January 26, 2011 Physics for Scientists&Engineers 1 12 12 Illustration Plot maximum height H as a function of the launch angle (for a given v0) 2 v0 H − y0 = sin 2 θ 0 2g January 26, 2011 Physics for Scientists&Engineers 1 13 13 Starting point: trajectory y = y0 + x tan θ 0 − g x2 2 2 v0 cos2 θ 0 January 26, 2011 Physics for Scientists&Engineers 1 14 14 Range The range, R, of a projectile is defined as the horizontal distance between the launching point and the point where the projectile reaches the same height again from which it started, y(R)=y0: y0 = y0 + R tan θ 0 − R tan θ 0 = tan θ 0 = g R2 2 2v0 cos 2 θ 0 g R2 2 2 2v0 cos θ 0 g R 2 2 2v0 cos θ 0 2 2 2v0 v0 R= sin θ 0 cos θ 0 = sin 2θ 0 g g January 26, 2011 Physics for Scientists&Engineers 1 15 15 Maximum Range For which angle is the range at maximum (for a given v0)? Take derivative with of R respect to angle: 2 2 dR d v0 v0 = sin 2θ 0 = 2 cos 2θ 0 dθ 0 dθ 0 g g This is 0 when cos 2θ 0 = 0 In other words, the launch angle has to be 45 degrees. Maximum range for an ideal projectile is then: 2 v0 = g 16 Rmax January 26, 2011 Physics for Scientists&Engineers 1 16 Illustration Plot range as a function of launch angle January 26, 2011 Physics for Scientists&Engineers 1 17 17 Throwing a Baseball Problem • What is the maximum height that a baseball reaches if it is thrown from second base to first base and from third base to first base, released from a height of 6.0 ft, with a speed of 90 mph, and caught at the same height? THINK • • • • • The baseball diamond is a square with 90 ft sides The distance from second base to first base is 27.4 m The distance from third base to first base is 38.3 m The speed of the ball is 40.2 m/s We will equate the distances between the bases with the range Physics for Scientists&Engineers 1 18 January 26, 2011 18 Throwing a Baseball SKETCH January 26, 2011 Physics for Scientists&Engineers 1 19 19 Throwing a Baseball (2) RESEARCH • To obtain the initial launch angle of the ball we use our equation for the range 2 v0 1 d R = d12 = sin 2θ0 θ0 = sin−1 12 2 g 2 v0 • The maximum height is given by The 2 v0 sin2 θ0 H = y0 + 2g SIMPLIFY 2 v0 sin2 H = y0 + January 26, 2011 1 −1 d12 sin 2 2 v0 2g Physics for Scientists&Engineers 1 20 20 Throwing a Baseball (3) CALCULATE 2 v0 sin2 H = y0 + 1 −1 d12 g sin 2 2 v0 2g 2 (40.2 m/s ) sin2 H = 1.83 m + 2 9.81 m/s2 ( (27.4 m )(9.81 m/s2 ) (40.2 m/s )2 ) H = 2.40367 m for second to first H = 2.95854 m for third to first ROUND H = 2.40 m for second to first H = 2.96 m for third to first January 26, 2011 Physics for Scientists&Engineers 1 21 21 Throwing a Baseball (4) DOUBLE-CHECK • It makes sense that a throw from third to first would have to be higher than a throw from second to first • The throw from second to first had to reach a height of 0.56 m (1.8 ft) above a straight line • The throw from third to first had to reach 1.03 m (3.4 ft) ft) above a straight line January 26, 2011 Physics for Scientists&Engineers 1 22 22 Clicker Quiz In the previous example, the third baseman threw the ball from third to first with an initial speed 40.2 m/s at an angle of 6.81°. The third baseman could throw the ball from third base to first base with the same initial speed but at a different angle. What What is that angle? A. B. C. D. E. 3.4° 13.6° 45.0° 51.8° 83.2° 2 v0 R = sin 2θ 0 g sin (θ ) = sin (180° − θ ) 2 (6.81° ) = 180° − 2θ 0 6.81° = 90° − θ 0 θ 0 = 83.2° January 26, 2011 Physics for Scientists&Engineers 1 23 23 Hang Time Problem • When a football team is forced to punt the ball away to the opponent, it is very important to kick the ball as far as possible but also to attain a sufficiently long hang time—that is, the ball should remain in the air long enough that the punt-coverage team has time to run down field and tackle the receiver right after the catch. • What are the initial angle and speed with which a football has to be punted so that its hang time is 4.41 s and it travels a distance of 49.8 m (= 54.5 yd)? THINK THINK • A punt is a special case of projectile motion for which the initial and final values of the vertical coordinate are both zero. If we know the range of the projectile, we can figure out the hang time from the fact that the horizontal component of the velocity vector remains at a constant value; thus, the hang time must simply be the range divided by this horizontal component of the velocity vector. The equations for hang time and range give us two equations in the two unknown quantities, v0 and θ0, that we are looking for. January 26, 2011 Physics for Scientists&Engineers 1 24 24 Hang Time (2) SKETCH January 26, 2011 Physics for Scientists&Engineers 1 25 25 Hang Time (3) RESEARCH • The range is given by 2 v0 R = sin 2θ0 g • The hang time is given by t= R v0 cosθ0 SIMPLIFY • Solve both equations for v02 and set them equal to each other 2 v0 R = sin 2θ0 g 2 v0 gR ց sin 2θ0 R2 ր 2 2 t cos θ0 R t= v0 cos θ0 January 26, 2011 v 2 0 gR R2 =2 sin 2θ0 t cos2 θ0 Physics for Scientists&Engineers 1 26 26 Hang Time (4) • Solve for θ0 using the trig identity sin2θ0 = 2sinθ0cosθ0 gR R2 =2 2 sin θ0 cosθ0 t cos2 θ0 gt 2 tan θ0 = 2R θ0 = tan −1 gt 2 2R • We can get the initial velocity from t= R v0 cosθ0 v0 = R t cosθ0 2 CALCULATE θ0 = tan−1 v0 = January 26, 2011 (9.81 m/s )(4.41 s) 2 2 (49.8 m ) = 62.4331° 49.8 m = 24.2013 m/s (4.41 s )cos (62.4331°) Physics for Scientists&Engineers 1 27 27 Hang Time (5) ROUND θ0 = 62.4° v0 = 24.4 m/s DOUBLE-CHECK • We know that the maximum range is reached with a launch launch angle of 45°. The punted ball here is launched at an initial angle that is significantly steeper, at 62.4°. Thus, the ball does not travel as far as it could go with the value of the initial speed that we computed. Instead, it travels higher and thus increases the hang time. January 26, 2011 Physics for Scientists&Engineers 1 28 28 ...
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