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Unformatted text preview: R e v ie w : Y e s t e r d a y
! Projectile trajectory
2 vy 0 Physics for Scientists & Engineers 1
Fall Semester 2006 Lecture 10 ! Maximum height H = y0 +
2 2g reached at xH = 2 v0 sin 2! 0 2g ! Range: R = v0 sin 2! 0 = 2 xH
g
2 v0 ! Maximum range: R = g ! Constant velocity along x
September 8, 2006 Physics for Scientists&Engineers 1 1 September 8, 2006 Physics for Scientists&Engineers 1 2 E x a m p le B a s e b a ll: B a t t in g
Question: ! If ball comes off bat with launch angle of 35° and with initial speed of 110 mph, how far will it ﬂy? How long will it be in the air? What will be its speed at the top of its trajectory? What will be its speed when it lands? Answer: (neglect air resistance for now, return to this later) ! Convert to SI: v0 = 110 mph = 49.2 m/s. ! Range: R = ! Air time:
September 8, 2006 E x a m p le B a s e b a ll: B a t t in g ( 2 )
Answer (cont.): ! At the top of the trajectory velocity has only a horizontal component: vx = vx 0 = v0 cos! 0 = 40.3 m/s ! Speed when baseball lands is the same as the speed when it left the bat, here 49.2 m/s
2 • Remember that in general v = v0 ! 2 g ( y ! y0 ) • And since y=y0 when the ball lands, we get v=v0. 2 v0 ( 49.2 m/s)2 sin 2! 0 = sin( 70°) = 231.5 m = 760 ft g 9.81 m/s 2 R 231.5 m t= = = 5.74 s v0 cos! 0 (49.2 m) " cos(35°)
Physics for Scientists&Engineers 1 3 September 8, 2006 Physics for Scientists&Engineers 1 4 “S h o o t ” t h e M o n k e y
! Where should one aim, if one want to hit a monkey, provided the monkey releases his grip and falls as soon as he hears the gun? E x a m p le : S h o o t t h e M o n k e y
! Monkey was released from a height of 2.5 m, ball was launched from height of 1 m. The distance between ball launcher and monkey is 4 m. ! Question: if the ball is launched with an initial speed of 8 m/s, at which height will the monkey be hit? 4m !
1m
September 8, 2006 Physics for Scientists&Engineers 1 5 September 8, 2006 Physics for Scientists&Engineers 1 2.5 m 6 E x a m p le : S h o o t t h e M o n k e y ( 2 )
Answer: ! First we calculate how long the ball needs to come to the monkey. They are 4 m apart, and the ball moves with 8 m/s => it takes 0.5 s for ball and monkey to meet (the fact that they both freefall during that time is not relevant!)
! Second, we can use y=y0gt2/2 for the position of the falling monkey. ! We get: 4m R e a lis t ic P r o je c t ile M o t io n
! As far as ideal projectile motion is concerned, beach balls and baseballs have the same trajectory ! Air resistance: drag force, proportional to v2 ! Leads to “ballistic curves” ! Example: baseballs launched at 35° and with 90 or 110 mph y = 2.5 m9.81·0.52/2 m = 1.27 m
1m
September 8, 2006 Physics for Scientists&Engineers 1 ! 2.5 m 7 September 8, 2006 Physics for Scientists&Engineers 1 8 R e a lis t ic P r o je c t ile M o t io n ( 2 )
! Spin of a projectile
• Spin of a thrown football contributes to stability • Topspin in tennis makes balls drop faster, backspin slower than expected from ideal projectile motion • Incidentally: a “rising fastball” does not really rise; however, it is thrown with extreme backspin so that it falls much slower than expected • Previous slide: curves were calculated for a realistic value of initial backspin of 2,000 rpm • Sidespin causes slices and hooks in golf, curveballs in baseball • Cause of trajectory changes due to spin: air molecules bouncing differently off surface of spinning projectile
September 8, 2006 Physics for Scientists&Engineers 1 9 September 8, 2006 P r o b le m ( 1 ) Portion 1: motion along a straight line with constant acceleration (xaxis = inclined plane) a = g ' si n !
!
A c q u ir e s v e lo c it y a lo n g t h e p la n e ! ! v0 Portion 2: ideal projectile motion ! ˆ a = ! g ' ey x0 = 10 y0 = 8 v0 x = v0 cos ! v0 y = v0 sin !
10 Physics for Scientists&Engineers 1 P r o b le m ( 2 )
! Motion along a straight line a = g ' si n ! l P r o b le m ( 3 )
! Ideal projectile motion
) " vy0 & g ' x2 $$ 0 y = + y0 ! x0 # ' ! 2 + vx 0 ( 2 vx 0 $ * %$ with and , ) "v & g ' x $ y0 $ 0 . ++# '+ 2 . + % vx 0 ( 2 vx 0  *$ $ , g' x2 .x! 2 . 2 vx 0  tan ! = n n p {} {} {} !
p ! v0 2 " v 2 = v0 + 2a ( x " x0 ) " vx 0 = v0 cos/ 0 v y 0 = v0 sin / 0 Remember /0 < 0 l= p cos ! p = 2g ' n cos ! v = 2 al = 2 g ' s i n ! Second Part v0 = 2 g ' n
September 8, 2006 Physics for Scientists&Engineers 1 2 ) g ' x0 , ) g ' x0 , g' y = + y0 ! x0 tan / 0 ! + + tan / 0 + x! x2 2 2. 2. 2 2 cos / 0 v0  * 2 cos 2 / 0 v0 2 cos 2 / 0 v0 * ! (positive) value of xx0 such that y(x) = 0?
g 2 => Quadratic equation in x to solve y = y0 + x tan ! 0 " 2 v 2 cos 2 ! x
0 0
11 September 8, 2006 Physics for Scientists&Engineers 1 12 ...
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