If a projectile is shot at a velocity of #12 m/s# and an angle of #pi/3#, how far will the projectile travel before landing?

Question

1234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T08:54:22

The distance is #=12.7m#

We need 2 equations

#sin2theta=2sinthetacostheta#

#x(t)=v_xt=u_0cos theta t#

and

#y(t)=v_yt-1/2g t^2#

#y(t)=u_0sintheta t-1/2g t^2#

In our case,

#y(t)=0#

So,

#1/2g t^2-u_0sintheta t =0#

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

The value #t=0# is when the projectile is shot

#t=(2u_0sintheta)/g#

Therefore,

#x(t)=u_0costheta *2u_0sintheta/g#

#=(u_0^2sin2theta)/g#

#=(12*12*sin(2/3pi))/9.8#

#=12.7 m#

graph{-0.136x^2+1.732x [-3.41, 16.59, -1.195, 8.805]}