A projectile is shot from the ground at an angle of #pi/4 # and a speed of #1/3 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 · 2017-04-15T06:34:32

The distance is #=0.003m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1/3*sin(1/4pi)#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=1/4sin(1/4pi)-g*t#

#t=1/4*1/g*sin(1/4pi)#

#=0.018s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=1/4cos(1/4pi)*0.018#

#=0.003m#