A projectile is shot from the ground at an angle of #pi/4 # and a speed of #36 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 · 2017-05-05T09:29:30

The distance is #=43.4m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=36*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=36sin(1/4pi)-g*t#

#t=36/g*sin(1/4pi)#

#=2.6s#

The greatest height is

#h=(36sin(pi/4))^2/(2g)=33.1m#

Resolving in the horizontal direction #rarr^+#

To find the horizontal distance, we apply the equation of motion

#s=u_x*t#

#=36cos(1/4pi)*2.6#

#=28.1m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(33.1^2+28.1^2)#

#=43.4m#