Question

If the ball leaves the projectile launcher at a speed of 2.4 m/s at an angle of 30ᴼ, and the projectile launcher is on a table at a height of 1.1 m, how far horizontally will it hit the ground? Give your answer in meters. An error margin of +- 0.01 m/s will be considered correct.

Answer 1

initial speed, u = 2.4m/s. angle with the horizontal, D=30° 4 horizontal component of the initial velocito u uloso -2.4 Los 60) = 2.0785/ vertical component of the initial velocity, lly usino = 2.4 Sin(o) = 1.2 m/s Cupward direction is assumed positive) vertical component of accelerations ay=-g=9,81aufs horizutal component of accelerations ax = 0 of the total time of flight is t, then h = lyt + axt2 =) –141=(1,2)- £49.81Xť? :> (4.90572-1.27 -1.1) = 0 solutioni- t = 0.6114s, -0,36 68 sec negative solution of time t is not acceptable. Hence; total time of flight, t = 0.6114 sec , horizontal o distance travelled, a= Uxt + 2 foxt 2 -(200785*0.61144) =11.2705m (Aus)