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#