A projectile is shot from the ground at an angle of #(5 pi)/12 # 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 · 2017-01-12T14:43:18

The distance #=0.14m#

We start by calcating the time to reach the maximum height

computing in the vertical direction #uarr^+#

#u=7/3sin((5/12)pi)#

#v=0#

#a=-g=-9.8#

We use the equation

#v=u+at#

#0=7/3sin((5/12)pi)-9.8t#

#9.8t=7/3sin((5/12)pi)#

#t=1/9.8*7/3sin((5/12)pi)=0.23s#

Computing in the horizontal direction #rarr^+#

#u=7/3cos((5/12)pi)#

#d=# distance from starting point

#d=7/3cos((5/12)pi)*0.23=0.14m#