A projectile is shot from the ground at an angle of #pi/4 # and a speed of #3/2 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 · 2017-05-31T10:13:23

The distance is #=0.13m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=3/2*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=3/2sin(1/4pi)-g*t#

#t=3/(2g)*sin(1/4pi)#

#=0.108s#

The greatest height is

#h=(3/2sin(1/4pi))^2/(2g)=0.057m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=3/2cos(1/4pi)*0.108#

#=0.115m#

The distance from the starting point is

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

#=sqrt(0.057^2+0.115^2)#

#=0.13m#