A projectile is shot at a velocity of #1 m/s# and an angle of #pi/12 #. What is the projectile's maximum height?

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Answer 1 · 2017-03-25T09:56:07

The maximum height is #=0.00358m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=1sin(1/12pi)#

#a=-g#

#v=0# at the maximum height

We apply the equation

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

#0=sin^2(pi/12)-2gh#

#h=1/(2g)*sin^2(pi/12)#

#=0.00358m#