A projectile is shot at an angle of #pi/6 # and a velocity of # 27 m/s#. How far away will the projectile land?

1 Answer
Feb 2, 2016

#x_max:64,354 meters#

Explanation:

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#v_o:#initial velocity
#theta:# angle
#v_o=27 m/s#
#R=pi/6 rad#
#t:# traveling time from A to B

#step 1:#
#theta/180=R/pi#
#theta/180=pi/(6.pi)#
#theta/180=1/6#
#theta=180/6=30^o#

#step 2:#
#g=9.81 m/s^2#
#v_x=v_o cos theta# ( no change during projectile)
#v_y=v_0 sin theta - g t#
#v_y=0# (at maximum height)
#0=v_0 sin theta - g t_m#
#g t_m=v_0 sin theta#
#t_m=(v_0 sin theta)/g#(time to maximum height)
#t=2.t_m#(traveling time from A to B)
#t=(2.v_o sin theta)/g#

#step 3:#
#x_max=v_x.t#
#x_max=v_o cos theta. 2(v_0 sin theta)/g#
#x_max=v_0^2 . 2.sin theta .cos theta/g#
#2. sin theta . cos theta=sin 2 theta#(trigonometric equation)
#x_max=v_o^2 .sin (2 theta)/g#
#theta=30^o #
#2 .theta=60^o#
#x_max=27^2 . sin 60/(9,18)#
#sin 30=0,866#
#x_max=(729 . 0,866)/(9,86)#
#x_max=(631,314)/(9,86)#
#x_max=64,354# meters