A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #3 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2017-01-11T08:12:24

The distance is #=0.3975m#

(1) The vertical motion #uarr^+#

#u=3sin(pi/3)#

#a=-9.8#

#v=0#

#s=h#

#v^2=u^2+2as#

#=>#, #0=(3sin(pi/3))^2-2*9.8*h#

#h=(3sin(pi/3))^2/(2*9.8)=0.3439m#

(2) Time to reach greatest height

#v=u+at#

#v=u+g t#

#0=3sin(pi/3)-9.8*t#

#t=(3sin(pi/3))/9.8=0.265s#

(3) The horozontal motion #rarr^+#

#u=3cos(pi/3)#

#d=ut=3cos(pi/3)*0.265=0.3975m#