A projectile is shot at an angle of #pi/3 # and a velocity of # 8 m/s#. How far away will the projectile land?

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Answer 1 · 2016-04-19T15:41:58

#x_max=x=5.655676106" "#meters

Solve for the time first

#y=v_0 sin theta*t+1/2*g*t^2#

Assuming level ground then #y=0#. Going up then going down then #y=0#

#0=8 sin (pi/3)t+1/2*(-9.8)*t^2#

#8*sqrt(3)/2*t-4.9*t^2=0#

#t(4sqrt3-4.9*t)=0#

#t_1=0" "#seconds and #t_2=(4sqrt3)/4.9=1.413919027" "#seconds

We can solve for the range #x# now

#x=v_0 cos theta * t_2#

#x=8* cos (pi/3) * 1.413919027#

#x=5.655676106" "#meters

God bless....I hope the explanation is useful.