# If a projectile is shot at a velocity of #52 m/s# and an angle of #pi/4#, how far will the projectile travel before landing?

##### 1 Answer

#### Explanation:

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=,52sin(pi/4)=26sqrt(2),ms^-1),(v=,"not required",ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #

So we can calculate

# :. 0 = 26sqrt(2)T + 1/2(-g)T^2#

# :. 1/2gT^2-26sqrt(2)T=0 #

# :. T((gT)/2-26sqrt(2)) =0#

# :. T =0, (52sqrt(2))/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 = 52cos(pi/4)T #

# :. s = (26sqrt(2))((52sqrt(2))/g) #

# :. s = 2704/g#

So using

#s=2704/9.8 = 275.918...= 276 m#