PROJECTILES Launched at an Anle
PROJECTILE MOTION AT AN ANGLE An bject launched int space withut mtie pwer f its wn is called a prjectile. If we nelect air resistance, the nly frce actin n a prjectile is its weiht, which causes its path t deiate frm a straiht line.
PROJECTILE MOTION AT AN ANGLE The prjectile has a cnstant hrizntal elcity and a ertical elcity that chanes unifrmly under the influence f the acceleratin due t raity.
PROJECTILE MOTION AT AN ANGLE The mre eneral case f prjectile mtin ccurs when the prjectile is fired at an anle.
PROJECTILE MOTION AT AN ANGLE The ertical elcity f the prjectile is the ertical cmpnent f the initial elcity, y = sin θ. The hrizntal elcity f the prjectile is the hrizntal cmpnent f the initial elcity, x = cs θ.
Prblem Slutin Stratey 1. Upward directin is psitie. The acceleratin due t raity () is dwnward thus makin it neatie. = - 9.8 m/s. Resle the initial elcity int its x and y cmpnents: x = cs θ y = sin θ This is ne f the mst imprtant steps that yu must d in rder t sle prblems f prjectiles launched at an anle. 3. The hrizntal and ertical cmpnents f its psitin at any instant is ien by: x = x t y = y t +½t
Prblem Slutin Stratey 4. The hrizntal and ertical cmpnents f its elcity at any ien instant are ien by: x = x y = y + t Where is a -9.8 m/s. 5. The final psitin and elcity can then be btained frm their cmpnents.
1. An artillery shell is fired with an initial elcity f 100 m/s at an anle f 30 abe the hrizntal. Find: a. Its psitin and elcity after 8 s. Gien: = 100 m/s, 30 t = 8 s = - 9.8 m/s x = x t = 86.6 m/s x 8 s = 69.8 m y = y t + ½ t x = 100 m/s cs 30 = 86.6 m/s y = 100 m/s sin 30 = 50 m/s x = x = 86.6 m/s y = y + t = 50 m/s + (-9.8 m/s )(8 s) = - 8.4 m/s = 50 m/s x 8 s + ½ x -9.8 m/s x (8 s) = 86.4 m
b. The time required t reach its maximum heiht. At the tp f the path: y = 0 y = y + t t y 50 m/s 9.8 m/s = 5.1 s c. The hrizntal distance (rane). Ttal time T = t = (5.1 s) = 10. s x = x t = 86.6 m/s x 10. s = 883.7 m
. A lf ball is hit and leaes the tee with a elcity f 5.0 m/s at 35.0 with the hrizn. a. What is the hrizntal displacement f the ball? Gien: V = 5.0 m/s θ = 35.0 V x = V cs θ V x = 5.0 m/s x cs 35.0 V x = 0.479 m/s V y = V sin θ V y = 5.0 m/s x sin 35.0 V y = 14.339 m/s t y 14.339 m/s t 9.8 m/s t.96 s x = V x t x = 0.479 m/s x.96 s x = 59.9 m
y b. Hw hih did the ball fly? Because the fliht is symmetrical, ne-half the time f the fliht is when the bject will be at its maximum heiht. Yu can use ne f the fllwin equatins. y 1/( t 1/ t )t t =.96 s, s.5t =.96/ = 1.463 s r y y 14.339 m/s 1.463 s y y t 1/ t 0.981 m y 10.493 m y 1/( y y 10.489 m y - 10.488 m )t.5-9.8 m/s.5(0 m/s 14.339 m/s)1.463 s (1.463 s)
Or yu can use the max heiht frmula, which is deried frm the fllwin equatin, where y = 0.0 m/s. y y y, slin fr y, with y 0.0 m/s, y y y y (14.339 m/s) 9.8 m/s 10.490 m
4. Find the anle f eleatin f a un that fires a shell with muzzle elcity f 15 m/s and hits a taret n the same leel but 1.55 km distant. Gien: = 15 m/s x = 1.55 km V V x y V V cs sin Usin the frmulas : x x at the tp r peak f and - 9.8 m/s 0 fy y x y t t t, t, hweer, t, ettinthe equatin t, sle fr t, y y y, but thisishalf the time,smultiplyby S ubint the first equatin. the fliht,the fy 0 m/s,, leainthe equatin as,,
x V x x y, but sub in the cs and sinfrmula, V x cs V sin V The tri functin,(sin cs ) sin x sin -1 9.8 m/s 1550 m 15 m/s sin (sin cs ) sin, sle fr, -1 x, θ = sin -1 (0.971) θ =76.4 θ = 38.
5. A plastic ball that is released with a elcity f 15 m/s stays in the air fr.0 s. a. At what anle with respect t the hrizntal was it released? = 15 m/s t = s time t maximum heiht = 1 s at the tp y = 0 y = y + t y 0 Rearrane : sin sin, ssubstitutinbth abe, sin sin t t t sin 9.8 m/s 1s) 15 m/s 1 = 40.8º
b. What was the maximum heiht achieed by the ball? y = y t +½t = (15)(sin 40.8º)(1) + ½(-9.8)(1) = 4.9 m y y y r max heiht max max max ( sin) () 4.901m y (15 m/s sin40.8 ( 9.8 m/s ) )
x t 6. An arrw was sht at an anle f 55º with respect t the hrizntal. The arrw landed at a hrizntal distance f 875 m. Find the elcity f the arrw at the tp f its path. Gien: θ = 55º x = 875 m x and x hrizntal x cs Yu will need t find the Ttal fliht time. x x cs t, s y = y t +½t if y = 0 0 = t ( y + ½ t) y = ½ t sin θ =½ t S ubfr time,t, frm the distance hrizntalfrmula. sin sin ½ t, t x sin cs x cs x, sle fr cs,
substitutin the time and slin fr x sin cs (sin 55 (9.8)(875) )(cs55 ) = 95.5 m/s At the tp f its path the arrw has y = and x = x x = 95.5 cs 55º x = 54.8 m/s
7. Find the rane f a un which fires a shell with muzzle elcity at an anle θ. What is the maximum rane pssible? t At tp y = 0 y = y + t 0 = sin θ t t = sin θ sin, thisis1/ the ttal fliht time. Ttal time = t x = x t x cs (sin cs ) sin
sin θ cs θ= ½ sin θ x (sin cs ) x sin Maximum rane is 45 since θ = 90
Assinment Ch 3, Pae 8-85 #1, 15, 19, 7, 8, 30, 35, 70. Extra Credit 47, and 51. D NOT just ie me the answer. Shw me HOW yu arried at the answer.