A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 8 m/s#. How far away will the projectile land?

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Answer 1 · 2016-11-13T15:39:37

Distance #=3.2m #

Initial velocity #u=8 ms^(-1)# at #theta=(5pi)/12#

We are going to use 2 equations

#h=utsintheta-g t^2/2#

Where #h=# height

Here #h=0#

#0=utsintheta-g t^2 /2=t(usintheta-(g t)/2)#

#t=2usintheta/g#

The second equation is

#d=utcostheta#

#d=ucostheta*(2usintheta)/g#

#sin(2theta)=2sinthetacostheta#

#d=u^2sin(2theta)/g#

#sin(5pi/6)=1/2#

Therefore #d=8^2sin((5pi)/6)/g=64*1/2*1/g#

#=32/10=3.2 m#

graph{3.73x-1.17x^2 [-2.06, 4.098, -0.144, 2.936]}