A projectile is shot from the ground at an angle of #pi/8 # and a speed of #7 /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 · 2017-02-02T11:36:15

The horizontal distance is #=0.2m#

Solving in the vertical direction #uarr^+#

#u=7/3sin(pi/8)ms^-1#

#v=0m s^-1#

#a=-g ms^-2#

#v=u+at#

#0=7/3sin(pi/8)-g t #

#t=(7/3sin(pi/8))/g#

This is the time to reach the greatest height

Solving in the horizontal direction #rarr^+#

#u=7/3cos(pi/8)#

#s=u*t=(7/3sin(pi/8))/g*7/3cos(pi/8)#

#=49/(9g)*sin(pi/8)cos(pi/8)#

#=0.2m#