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

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Answer 1 · 2016-12-09T16:14:58

The height is #=0.05m#

At the maximum height, the vertical component of the velocity is #=0#

The vertical component is #v=u_0sintheta#

#u_0=12ms^(-1)#

#theta=pi/12#

We use the equation

#v^2=u_0^2sin^2theta-2*g*h#

#0=u_0^2sin^2theta-2*g*h#

#h=(u_0^2sin^2(pi/12))/(2g)#

#=144*sin^2(pi/12)/(2*9.8)#

#=0.05m#