A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 4 m/s#. How far away will the projectile land?

As the projectile is launched at an angle, we will need to break the given velocity up into its #x#- and #y#-components. A simple diagram in the form of a right triangle will help.

Using basic trigonometry, we can see that

#v_x=vcos(theta)#

#v_y=vsin(theta)#

We will calculate both of these components, then use the #y# component of the velocity to solve for the total flight time of the projectile. Finally, we will use this value and the #x# component of the velocity to solve for the horizontal range of the projectile.

#v_x=4cos((5pi)/12)~~1.04m/s#

#v_y=4sin((5pi)/12)~~3.86m/s#

Assuming the launch and landing points of the projectile are at the same altitude, the total flight time of the projectile is twice the flight time between the launch and the maximum altitude of the projectile. We can find #Deltat# at the maximum altitude (#v_f=0#) and multiply by two, using #v_i=3.86m/s# and #a=-g=-9.8m/s^2#

This kinematic will do the trick: #v_(fy)=v_(iy)+a_yDeltat#. Rearrange and solve for #Deltat#.

#=>Deltat=(v_(fy)-v_(iy))/a_y#

#=>Deltat=(-v_(iy))/-g#

#=>Deltat=(-3.86m/s)/(-9.8m/s^2)#

#=>Deltat=0.394s#

The total flight time is then #2*0.394=0.788s#.

Now we can use this kinematic to find the range of the projectile:

#x_f=x_i+v_(ix)Deltat*1/2a_xDeltat^2#

Where #x_i=0m#, #v_(ix)=1.04m/s#, #Deltat=0.788s#, and #a_x=0# (no horizontal acceleration in simple projectile motion). We solve for #x_f#.

#=>x_f=(1.04m/s)(0.788s)=0.82m#

Hope that helps!