A projectile is shot from the ground at an angle of #pi/8 # and a speed of #8 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

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Answer 1 · 2017-05-11T11:53:46

The distance is #=2.34m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=8*sin(1/8pi)#

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=8sin(1/8pi)-g*t#

#t=8/g*sin(1/8pi)#

#=0.31s#

The greatest height is

#h=(8sin(pi/8))^2/(2g)=0.48m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=8cos(1/8pi)*0.31#

#=2.29m#

The distance from the starting point is

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

#=sqrt(0.48^2+2.29^2)#

#=2.34m#