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
Dec 6, 2016

#2.55 m#

Explanation:

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

#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #

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Vertical Motion
Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile is in the air be #T#, then then total vertical displacement will be #0#, we would expect two solutions, one will be #t=0#

# { (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 #T# using #s=ut+1/2at^2#

# :. 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, #t =T#

# s=(5)cos(pi/4)T #
# :. s=(5sqrt(2))/2T #
# :. s=(5sqrt(2))/2 * (5sqrt(2))/g #
# :. s=25/g #

So using #g=9.8 ms^-2# we have.

#s=25/9.8 = 2.55102...= 2.55 m#