If a projectile is shot at a velocity of #5 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 is in the air be
# { (s=,0,m),(u=,(5)sin(pi/4)=5/2sqrt(2),ms^-1),(v=,"not required",ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #
So we can calculate
# :. 0=5/2sqrt(2)T-1/2gT^2 #
# :. 1/2T(5sqrt(2)-gT)=0 #
# :. T=0, (5sqrt(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,
# s=(5)cos(pi/4)T #
# :. s=(5sqrt(2))/2T #
# :. s=(5sqrt(2))/2 * (5sqrt(2))/g #
# :. s=25/g #
So using
#s=25/9.8 = 2.55102...= 2.55 m#