A projectile is shot at a velocity of # 4 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

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Answer 1 · 2017-03-29T09:35:31

The peak height is #=0.204m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=4sin(1/6pi)#

#a=-g#

#v=0# at the maximum height

We apply the equation

#v^2=u^2+2ah#

#0=16sin^2(pi/6)-2gh#

#h=1/(2g)*16sin^2(pi/6)#

#=0.204m#