A projectile is shot from the ground at an angle of #pi/4 # and a speed of #2/5 ms^-1#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer
Dec 8, 2016

#8.1mm#

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 maximum point be #T#, at the point the velocity will momentary be zero.

# { (s=,"not required",m),(u=,2/5 sin(pi/4)=1/5sqrt(2),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #

So we can calculate #T# using #v=u+at#

# :. 0=1/5sqrt(2)-gT #
# :. gT=1/5sqrt(2) #
# :. T=sqrt(2)/(5g)#

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 #s# using #s=ut#

# s=2/5cos(pi/4)T #
# :. s=1/5sqrt(2)T #
# :. s=1/5sqrt(2)sqrt(2)/(5g) #
# :. s=2/(25g) #

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

#s=2/245 = 0.008163...= 0.0081 m#, or #8.1mm#