A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #7/4 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 · 2017-02-01T08:29:12

Horizontal distance #=0.135m#

Solving in the vertical direction #uarr^+#

#u=7/4sin(pi/3)ms^-1#

#v=0m s^-1#

#a=-g ms^-2#

#v=u+at#

#0=7/4sin(pi/3)-g t #

#t=(7/4sin(pi/3))/g#

This is the time to reach the greatest height

Solving in the horizontal direction #rarr^+#

#u=7/4cos(pi/3)#

#s=u*t=(7/4sin(pi/3))/g*7/4cos(pi/3)#

#=49/(16g)*sin(pi/3)cos(pi/3)#

#=0.135m#