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
Dec 14, 2016

#276 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 takes to its return to the ground after launch be #T#, We would expect two solutions as obviously T=0 is one solution. The total displacement will be zero.

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

# :. 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, #t =T#

So we can calculate the horizontal displacement #s# using #s=ut#

# s = 52cos(pi/4)T #
# :. s = (26sqrt(2))((52sqrt(2))/g) #
# :. s = 2704/g#

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

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