A projectile is shot at an angle of #pi/4 # and a velocity of # 6 ms^-1#. How far away will the projectile land?

1 Answer
Mar 25, 2018

# 3.67 \ m # (3 sf)

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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Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with #a=0#).

Suppose that the projectile will travel a distance #x \ m# in time #T \ s#, we must resolve the initial speed in the horizontal direction:

# { (s=,x,m), (u=,6 cos (pi/4)=3sqrt(2),ms^-1), (v=,"Not Required",ms^-1), (a=,0,ms^-2), (t=,T,s) :} #

So applying #s=ut+1/2at^2# we get

# x = 3sqrt(2)T#

Vertical Motion

The projectile travels under constant acceleration due to gravity. Its displacement will be #0# at the point where the projectile is at ground level, leading to two solution #t=0# (trivial) and #t=T#. Considering upwards as positive, and again resolving the initial speed (vertically this time):

# { (s=,0,m), (u=,6 sin (pi/4)=3sqrt(2),ms^-1), (v=,0,ms^-1), (a=,-g,ms^-2), (t=,T,s) :} #

Applying #s=ut+1/2at^2# we have:

# 0 = 3sqrt(2)T+1/2(-g)T^2 #
# :. 0 = 6sqrt(2)T-gT^2 #
# :. T(6sqrt(2)-gT) = 0 #

Leading to two possible solution:

# { (T = 0,,"(trivial solution)"),(6sqrt(2)-gT=0,,) :} #

Hence we have:

# 6sqrt(2)-gT = 0 => T = (6sqrt(2))/g #

And using the result we established from the horizontal motion, we have:

# x = 3sqrt(2) * (6sqrt(2))/g#
# \ \ = 36/g#

If we use #g=9.8 \ ms^(-2)# we get:

# x = 3.67 \ m # (3 sf)