A projectile is shot at a velocity of #9 m/s# and an angle of #pi/8 #. What is the projectile's peak height?

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Answer 1 · 2017-01-30T14:33:23

#"the answer is "h_p=0.60" m"#

#"the peak height of a projectile can be calculated using:"#

#h_p=(v_i^2.sin^2 alpha)/(2.g)#

#"where;#

#v_i:"initial velocity ,"v_i=9" m/s"#
#alpha=pi/8" , "sin alpha=0.38268343" , "sin^2 alpha=0.14644661#

#g=9.81 (N)/(kg)#

#h_p:"peak height"#

#h_p=(9^2* 0.14644661)/(2*9.81)#

#h_p=(81* 0.14644661)/(19.62)#

#h_p=0.60" m"#