A projectile is shot at a velocity of # 18 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

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Answer 1 · 2018-03-15T18:08:21

The peak height is #=4.13m#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=(18)*sin(1/6pi)ms^-1#

Applying the equation of motion

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

At the greatest height, #v=0ms^-1#

The acceleration due to gravity is #a=-g=-9.8ms^-2#

Therefore,

The greatest height is #h_y=s=(0-(18)sin(1/6pi))^2/(-2g)#

#h_y=(18sin(1/6pi))^2/(2*9.8)=4.13m#