If a projectile is shot at an angle of #pi/6# and at a velocity of #18 m/s#, when will it reach its maximum height??

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Answer 1 · 2016-03-30T13:04:13

Time of reaching at maximum height
#t =(usinalpha)/g=(18*sin(pi/6))/9.8=0.91s#

Answer 2 · 2016-03-30T13:04:36

#t~=0,92 " s"#

#g=9,82 " m/s^2#

#v_i=18" "m/s#

#alpha=pi/6#

#sin alpha=0,5#

#t=(v_i*sin alpha)/g#

#t=(18*0,5)/(9,81)#

#t=9/(9,81)#

#t~=0,92 " s"#