A projectile is shot at a velocity of #27 m/s# and an angle of #pi/12 #. What is the projectile's maximum height?

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Answer 1 · 2017-04-28T06:32:12

The maximum height is #=2.49m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=27*sin(1/12pi)#

At the maximum height, #v=0#

We apply the equation of motion

#v^2=u^2+2as#

to calculate the greatest height

#0=(27sin(1/12pi)^2-2g*h#

#h=1/(2g)*(27sin(1/12pi))^2#

#=2.49m#