A projectile is shot at a velocity of # 16 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

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Answer 1 · 2017-12-27T08:07:11

The peak height is #=3.27m#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u=16sin(1/6pi)ms^-1#

Let the peak height be #=hm#

At the greatest heignt #v=0ms^-1#

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

Applying the equation of motion

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

#h=(u^2-v^2)/(2g)=((16sin(1/6pi))^2-0)/(2g)=3.27m#