A projectile is shot at a velocity of #3 m/s# and an angle of #pi/8 #. What is the projectile's peak height?

1 Answer
ali ergin
Feb 3, 2016

#h_(peak)=0,00888# #"meters"#

Explanation:

#"the formula needed to solve this problem is:"#
#h_(peak)=(v_i^2*sin^2 theta/(2*g))#
#v_i=3 m/s#
#theta =180/cancel(pi)*cancel(pi)/8#
#theta =180/8#
#sin theta=0,13917310096#
#sin^2 theta=0,0193691520308#
#h_(peak)=3^2*(0,0193691520308)/(2*9,81)#
#h_(peak)=9*(0,0193691520308)/(19,62)#
#h_(peak)=0,00888# #"meters"#

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