Firing an Ideal Projectile

Transcription

1 92 Chapter 13: Vector-Valued Functions and Motion in Space

2 13.2 Modelin Projectile Motion 921 r at time t v v cos i a j (a) v sin j Newton s second law of motion sas that the force actin on the projectile is equal to the projectile s mass m times its acceleration, or msd 2 r> 2 d if r is the projectile s position vector and t is time. If the force is solel the ravitational force -mj, then We find r as a function of t b solvin the followin initial value problem. Differential equation: d 2 r 2 Initial conditions: r = r and dr The first interation ives m d 2 r 2 = -mj and d 2 r 2 = -j = -j. = v when t = (, ) v A second interation ives dr = -stdj + v. a j r =- 1 2 t 2 j + v t + r. r i j R Horizontal rane (b) Substitutin the values of and from Equations (1) and (2) ives v r r =- 1 2 t 2 j + s cos ai + s sin aj + ('''''')''''''* v t FIGURE 13.9 (a) Position, velocit, acceleration, and launch anle at t =. (b) Position, velocit, and acceleration at a later time t. Collectin terms, we have Ideal Projectile Motion Equation r = s cos ai + as sin a t 2 bj. (3) Equation (3) is the vector equation for ideal projectile motion. The anle a is the projectile s launch anle (firin anle, anle of elevation), and, as we said before, is the projectile s initial speed. The components of r ive the parametric equations = s cos a and = s sin a t 2, (4) where is the distance downrane and is the heiht of the projectile at time t Ú. EXAMPLE 1 Firin an Ideal Projectile A projectile is fired from the oriin over horizontal round at an initial speed of 5 m> sec and a launch anle of 6. Where will the projectile be 1 sec later?

3 922 Chapter 13: Vector-Valued Functions and Motion in Space Solution We use Equation (3) with = 5, a = 6, = 9.8, and t = 1 to find the projectile s components 1 sec after firin. r = s cos ai + as sin a t 2 bj = s5d a 1 23 bs1di + as5d a 2 2 b1 - a1 2 bs9.8ds1dbj L 25i + 384j. Ten seconds after firin, the projectile is about 384 m in the air and 25 m downrane. Heiht, Fliht Time, and Rane Equation (3) enables us to answer most questions about the ideal motion for a projectile fired from the oriin. The projectile reaches its hihest point when its vertical velocit component is zero, that is, when d For this value of t, the value of is = sin a - t =, or t = sin a. ma = s sin ad a sin a To find when the projectile lands when fired over horizontal round, we set the vertical component equal to zero in Equation (3) and solve for t. s sin a t 2 = b a sin a b = s sin ad 2. 2 t a sin a tb = Since is the time the projectile is fired, s2 sin ad> must be the time when the projectile strikes the round. To find the projectile s rane R, the distance from the oriin to the point of impact on horizontal round, we find the value of the horizontal component when t = s2 sin ad>. = s cos a R = s cos ad a 2 sin a b = 2 s2 sin a cos ad = 2 sin 2a The rane is larest when sin 2a = 1 or a = 45. t =, t = 2 sin a

4 13.2 Modelin Projectile Motion 923 Heiht, Fliht Time, and Rane for Ideal Projectile Motion For ideal projectile motion when an object is launched from the oriin over a horizontal surface with initial speed and launch anle a: Maimum heiht: ma = s sin ad 2 2 Fliht time: t = 2 sin a Rane: R = 2 sin 2a. EXAMPLE 2 Investiatin Ideal Projectile Motion Find the maimum heiht, fliht time, and rane of a projectile fired from the oriin over horizontal round at an initial speed of 5 m> sec and a launch anle of 6º (same projectile as Eample 1). Solution Maimum heiht: ma = s sin ad 2 2 Fliht time: t s5 sin 6 d2 = 2s9.8d = 2 sin a L 9566 m 2s5d sin 6 = 9.8 Rane: R = 2 sin 2a L 88.4 sec 1, , 15, 2, = s5d2 sin From Equation (3), the position vector of the projectile is r = s cos ai + as sin a t 2 bj L 22,92 m = s5 cos 6 i + as5 sin s9.8 2 bj FIGURE 13.1 The raph of the projectile described in Eample 2. = 25ti + AA2523Bt - 4.9t 2 Bj. A raph of the projectile s path is shown in Fiure Ideal Trajectories Are Parabolic It is often claimed that water from a hose traces a parabola in the air, but anone who looks closel enouh will see this is not so. The air slows the water down, and its forward proress is too slow at the end to keep pace with the rate at which it falls.

5 924 Chapter 13: Vector-Valued Functions and Motion in Space v (, ) What is reall bein claimed is that ideal projectiles move alon parabolas, and this we can see from Equations (4). If we substitute t = >s cos ad from the first equation into the second, we obtain the Cartesian-coordinate equation = - a 2 2 cos 2 a b 2 + stan ad. This equation has the form = a 2 + b, so its raph is a parabola. FIGURE The path of a projectile fired from s, d with an initial velocit v at an anle of a derees with the horizontal. Firin from s, d If we fire our ideal projectile from the point s, d instead of the oriin (Fiure 13.11), the position vector for the path of motion is r = s + s cos adi + a + s sin a - 1 (5) 2 t 2 bj, as ou are asked to show in Eercise 19. EXAMPLE 3 Firin a Flamin Arrow To open the 1992 Summer Olmpics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olmpic torch with a flamin arrow (Fiure 13.12). Suppose that Rebollo shot the arrow at a heiht of 6 ft above round level 9 ft from the 7-ft-hih cauldron, and he wanted the arrow to reach maimum heiht eactl 4 ft above the center of the cauldron (Fiure 13.12). Photoraph is not available. FIGURE Spanish archer Antonio Rebollo lihts the Olmpic torch in Barcelona with a flamin arrow. (a) Epress in terms of the initial speed and firin anle a. ma (b) Use ma = 74 ft (Fiure 13.13) and the result from part (a) to find the value of sin a. (c) Find the value of cos a. (d) Find the initial firin anle of the arrow.

6 13.2 Modelin Projectile Motion 925 Solution ma 74' 9' v 6' (a) We use a coordinate sstem in which the positive -ais lies alon the round toward the left (to match the second photoraph in Fiure 13.12) and the coordinates of the flamin arrow at t = are = and = 6 (Fiure 13.13). We have = + s sin a t 2 Equation (5), j-component NOT TO SCALE FIGURE Ideal path of the arrow that lit the Olmpic torch (Eample 3). = 6 + s sin a t 2. We find the time when the arrow reaches its hihest point b settin d> = and solvin for t, obtainin t = sin a. = 6 For this value of t, the value of is ma = 6 + s sin ad a sin a = 6 + s sin ad 2. 2 (b) Usin ma = 74 and = 32, we see from the preceedin equation in part (a) that 74 = 6 + s sin ad 2 2s32d b a sin a b or sin a = 2s68ds64d. (c) When the arrow reaches ma, the horizontal distance traveled to the center of the cauldron is = 9 ft. We substitute the time to reach ma from part (a) and the horizontal distance = 9 ft into the i-component of Equation (5) to obtain = + s cos a 9 = + s cos a = s cos ad a sin a b. Equation (5), i-component = 9, = t = s sin ad> Solvin this equation for cos a and usin = 32 and the result from part (b), we have cos a = (d) Parts (b) and (c) toether tell us that 9 sin a = s9ds32d 2s68ds64d. tan a = sin a cos a = A 2s68ds64dB2 s9ds32d = 68 45

7 926 Chapter 13: Vector-Valued Functions and Motion in Space or This is Rebollo s firin anle. a = tan -1 a b L Projectile Motion with Wind Gusts The net eample shows how to account for another force actin on a projectile. We also assume that the path of the baseball in Eample 4 lies in a vertical plane. EXAMPLE 4 Hittin a Baseball A baseball is hit when it is 3 ft above the round. It leaves the bat with initial speed of 152 ft> sec, makin an anle of 2 with the horizontal. At the instant the ball is hit, an instantaneous ust of wind blows in the horizontal direction directl opposite the direction the ball is takin toward the outfield, addin a component of -8.8i sft>secd to the ball s initial velocit s8.8 ft>sec = 6 mphd. (a) Find a vector equation (position vector) for the path of the baseball. (b) How hih does the baseball o, and when does it reach maimum heiht? (c) Assumin that the ball is not cauht, find its rane and fliht time. Solution (a) Usin Equation (1) and accountin for the ust of wind, the initial velocit of the baseball is The initial position is r Interation of d 2 r> 2 = i + 3j. = -j ives A second interation ives v = s cos adi + s sin adj - 8.8i = s152 cos 2 di + s152 sin 2 dj - s8.8di = s152 cos 2-8.8di + s152 sin 2 dj. dr = -stdj + v. r =- 1 2 t 2 j + v t + r. Substitutin the values of v and r into the last equation ives the position vector of the baseball. r =- 1 2 t 2 j + v t + r = -16t 2 j + s152 cos 2-8.8i + s152 sin 2 j + 3j = s152 cos 2-8.8i + A3 + (152 sin 2-16t 2 Bj.

8 13.2 Modelin Projectile Motion 927 (b) The baseball reaches its hihest point when the vertical component of velocit is zero, or Solvin for t we find d = 152 sin 2-32t =. t = 152 sin 2 32 L 1.62 sec. Substitutin this time into the vertical component for r ives the maimum heiht ma = 3 + s152 sin 2 ds1.62d - 16s1.62d 2 L 45.2 ft. That is, the maimum heiht of the baseball is about 45.2 ft, reached about 1.6 sec after leavin the bat. (c) To find when the baseball lands, we set the vertical component for r equal to and solve for t: 3 + s152 sin 2-16t 2 = 3 + s t 2 =. The solution values are about t = 3.3 sec and t = -.6 sec. Substitutin the positive time into the horizontal component for r, we find the rane R = s152 cos 2-8.8ds3.3d L 442 ft. Thus, the horizontal rane is about 442 ft, and the fliht time is about 3.3 sec. In Eercises 29 throuh 31, we consider projectile motion when there is air resistance slowin down the fliht.