Question

1. From the castle wall 20m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above the horizontal.

1. Find the initial velocity components.
2. Find the maximum height of the arrow.
3. Find the range.
4. Find an arrow final velocity.

Answer 1

(a) Let v_x be the horizontal component of velocity and v_y be the vertical component

           v_x = v.Cos(40^o)= 45*0.766 = 34.47 m/s

           v_y = v.Sin(40^o) = 45*0.64 = 28.92 m/s

(b) Maximum height of the arrow = Height of wall + Height attained by arrow from wall

                                                      = 20 +

                                                     = 20 +     => 20 +

                                                    = 20 + 42.67

                                                    = 62.67 m

                    

(c) Total Range = Range covered from initial to the point horizontal level to wall height(A to B)

                                  + Range from B to C

                       =    +

                      =    +           ________using Second Equation of motion

                      = 202.5 + 18.27

                     =    220.77 m

(d) For the final velocity

             The horizontal component of velocity remains same as initial = 34.47 m/s

           The vertical velocity will be 28.92 m/s + Velocity gained from B to C

                                           = 28.92 + 9.8*(0.53)      ____________using First equation of Motion

                                            = 34.11 m/s

        So the angle between horizontal and final velocity vector will be

                                              = =

                                             = 44.69^o