A projectile is projected with initial velocity #(6i+8j)# #m##/sec#. If #g=10m*s^-2#, what is the horizontal range?

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Answer 1 · 2017-09-28T15:14:54

#9.6m#

The question gives us the information in vector form so we can use the vector form of the constant acceleration equations

we have the following, for convenience we will change the vectors to column form

#vecu=(6veci+8vecj)ms^(-1)=((6),(8))ms^(-1)#

#vecg=((0),(-10))ms^(-2)#

the equation we will use is the vector equivalent to

#s=ut+1/2at^2#

#vecr=vecu t+1/2veca t^2#

#vecr=((r_x),(r_y))=((6),(8))t-1/2((0),(10))t^2#

for the horizontal range

#r_y=0#

#:.((r_x),(0))=t[((6),(8))-((0),(5))t]#

equating components

#y-"components"#

#0=t(8-5t)#

#=>t=0, "or "t=8/5sec#

for the range we take the non zero value

#x-"component"#

#r_x=3t=6xx8/5=48/5=9.6m#