A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #4/5 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?

1 Answer
Feb 24, 2018

The distance is #=0.034m#

Explanation:

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=(4/5)sin(5/12pi)ms^-1#

Applying the equation of motion

#v^2=u^2+2as#

At the greatest height, #v=0ms^-1#

The acceleration due to gravity is #a=-g=-9.8ms^-2#

Therefore,

The greatest height is #h_y=s=(0-(4/5)sin(5/12pi))^2/(-2g)#

#h_y=(4/5sin(5/12pi))^2/(2g)=0.03m#

The time to reach the greatest height is #=ts#

Applying the equation of motion

#v=u+at=u- g t #

The time is #t=(v-u)/(-g)=(0-(4/5)sin(5/12pi))/(-9.8)=0.079s#

Resolving in the horizontal direction #rarr^+#

The velocity is constant and #u_x=(4/5)cos(5/12pi)#

The distance travelled in the horizontal direction is

#s_x=u_x*t=(4/5)cos(5/12pi)*0.079=0.016m#

The distance from the starting point is

#d=sqrt(h_y^2+s_x^2)=sqrt(0.03^2+0.016^2)=0.034m#