A projectile is shot at an angle of #pi/8 # and a velocity of # 4 "m/s"#. How far away will the projectile land?

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Answer 1 · 2016-12-10T13:51:11

The distance is #=1.15m#

To find the range of the projectile, the height #h=0#

We use the equation

#h=u_0tsintheta-1/2g t^2#

#t(u_0sintheta-1/2 g t)=0#

#t=0# is the time when the projectile is shot

#t=(2u_0sintheta)/g#

The horizontal component of the velocity #=u_0costheta#

So, the range

#=(2u_0sintheta)/g*u_0costheta#

#=u_0^2sin(2theta)/g#

as, #2sinthetacostheta=sin2theta#

The range

#=u_0^2sin(2theta)/g=4^2*sin(pi/4)*1/9.8#

#=16*sqrt2/2*1/9.8=1.15m#