A projectile is shot at a velocity of # 64 m/s# and an angle of #pi/4 #. What is the projectile's peak height?

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Answer 1 · 2016-01-16T11:15:50

#104.49m#

We may consider the horizontal and vertical components separately.

For the y-direction, the acceleration is constant under gravity, hence the equations of motion for constant linear acceleration are valid.

#v^2=u^2+2ax#

#therefore x=(v^2-u^2)/(2a)#

#=(0^2-(64sin(pi/4))^2)/(2xx9.8)#

#=104.49m#.

Alternatively, you could also have used calculus by finding an equation for #x(t)# and setting its derivative to zero to find its maximum value.