A projectile is shot at a velocity of # 12 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

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Answer 1 · 2016-07-03T16:56:43

#h_m=1.83" "m#

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#"The projectile's peak height can be calculated using :"#

#h_m=(V_i^2*sin^2 alpha)/(2*g)#

#h_m" represents peak height"#

#v_i:"initial velocity"#

#v_i=12" "m/s#

#alpha=(pi/6)" ; "sin (pi/6)=1/2" ; "sin^2 (pi/6)=1/4#

#g=9.81" "m/s^2#

#h_m=(12^2*1/4)/(2*9.81)#

#h_m=36//19.62)#

#h_m=1.83" "m#