A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #5/8 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer
Apr 3, 2016

#v_x=0,16 " m/s"#
#v_y=0,60" m/s"#
#"x-component of velocity at any time:" v_x=v_i*cos alpha#
#"y-component of velocity at any time:"v_y=v_i*t*sin alpha-g*t #
#x_m=0,02" "m#
#t=0,06" s"#

Explanation:

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#"x-component of the initial velocity : "v_x=v_*cos alpha#

#v_x=5/8*cos ((5pi)/12)#

#v_x=0,16 " m/s"#

#"y-component of the initial velocity : "v_y=v_i*sin alpha#

#v_y=5/8*sin((5pi)/12)#

#v_y=0,60" m/s"#

#"x-component of velocity at any time:" v_x=v_i*cos alpha#
#"y-component of velocity at any time:"v_y=v_i*t*sin alpha-g*t #

#"for maximum x: "x_m=(v_i^2*sin2 alpha)/g#

#x_m=((5/8)^2*sin 2*(5 pi)/12)/(9,81)#

#x_m=(25*sin((5pi)/6))/(64*9,81)=(25*0,5)/(64*9,81)#

#x_m=0,02" "m#

#"for time to maximum height : "t=(v_i*sin alpha)/g#

#t=(5*sin((5pi)/12))/(8*9,81)=(4,8296291314)/(78,48)#

#t=0,06" s"#