# A projectile is shot at an angle of #pi/12 # and a velocity of #88 m/s#. How far away will the projectile land?

##### 1 Answer

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

**Vertical Motion**

Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile takes to its return to the ground after launch be

# { (s=,0,m),(u=,88 sin(pi/12),ms^-1),(v=,"not required",ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #

So we can calculate

# :. 0 = 88sin(pi/12)T + 1/2(-g)T^2#

# :. 1/2gT^2-88sin(pi/12)T=0 #

# :. T((gT)/2-88sin(pi/12)) =0#

# :. T =0, T = (176sin(pi/12))/g#

**Horizontal Motion**

Under constant speed (NB we can still use "suvat" equation with a=0). Thje projectile will be in in the air for the same time,

So we can calculate the horizontal displacement

# s = 88cos(pi/12)T #

# :. s = (88cos(pi/12)(176sin(pi/12)))/g #

# :. s = (176*44*2cos(pi/12)sin(pi/12))/g#

# :. s = (7744sin((2pi)/12))/g#

# :. s = ((7744)(0.5))/g#

# :. s = 3872/g#

So using

#s=3872/9.8 = 395.102...= 395 m#