A projectile is shot at a velocity of #1 5 m/s# and an angle of #pi/8 #. What is the projectile's peak height?

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Answer 1 · 2017-07-04T18:37:50

The peak height is #=1.68m#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_y=vsintheta=15*sin(1/8pi)#

The acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

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

to calculate the greatest height

#0=(15sin(1/8pi)^2)-2gh#

The greatest height is

#h=(15sin(1/8pi))^2/(2g)=1.68m#