Question

1:17 2:26 Exit 1: What should the initial velocity of an arrow fired at 50° upward from the horizon be if the arrow was released by an archer that stands 2 meters tall aiming to hit a practice target (bulls eye) that also stands 2 meters high located 0.5 kilo meter away. Enter answer... 2. 2: A particle is fired at a velocity of 20 meters per second in a direction making an angle of 25 upward with the horizon. a: What is the maximum height reached by the particle? b: What is the total flight time of the particle (between launch and touch down)? c: What is the horizontal range (Maximum X distance)? d: What is the magnitude of the velocity of the particle just before it hits the ground?

Answer 1

u Solution R= Range = horizontal distance itsinde R= using g u² =Rg (g=918 Sin20 mis2) i2m =) u = 4975.59 U=70153786 m/s 50" x=R=.5km = 5oom am 2 2 u R- g given that u=20m/s, o=25° (a) H= uhsino = 20 x sih laro zg 2x9.8 Bu TH= 3.645 m Time of flight T= zusino 1.7249 = 1.725 sec = Time of flight (T=1.725 sec R ut sinza = 31.267 m -> Hońzontal distance = 31.2677 g (d) velocity Just before touching will be laundung time C energy will not love so same Kinche' energy at ground) u= a= 20m/s = v=20 m/s magnéhåde same