A bullet leaves the muzzle of a gun at 250 m/s to hit a target 120m away at the level of the muzzle (1.9 m above ground), the gun must be aimed above the target. (Ignore air resistance.) How far above the target is that point?

1 Answer
Jun 12, 2015

The distance is #1.11m#.

Explanation:

To answer at the question we have to find the angle #alpha# above the horizontal, at which we have to shoot.

The motion is a parabolic motion, that is the composition of two motion:

the first, horizontal, is an uniform motion with law:

#x=x_0+v_(0x)t#

and the second is a decelerated motion with law:

#y=y_0+v_(0y)t+1/2g t^2#,

where:

  • #(x,y)# is the position at the time #t#;
  • #(x_0,y_0)# is the initial position;
  • #(v_(0x),v_(0y))# are the components of the initial velocity, that are, for the trigonometry laws:
    #v_(0x)=v_0cosalpha#
    #v_(0y)=v_0sinalpha#
    (#alpha# is the angle that the vector velocity forms with the horizontal);
  • #t# is time;
  • #g# is gravity acceleration.

To obtain the equation of the motion, a parabola, we have to solve the system between the two equation written above.

#x=x_0+v_(0x)t#
#y=y_0+v_(0y)t+1/2g t^2#.

Let's find #t# from the first equation and let's substitue in the second:

#t=(x-x_0)/v_(0x)#

#y=y_0+v_(0y)(x-x_0)/v_(0x)-1/2g*(x-x_0)^2/v_(0x)^2# or:

#y=y_0+v_0sinalpha(x-x_0)/(v_0cosalpha)-1/2g*(x-x_0)^2/(v_0^2cos^2alpha)# or

#y=y_0+sinalpha(x-x_0)/cosalpha-1/2g*(x-x_0)^2/(v_0^2cos^2alpha)#

To find the range we can assume:

#(x_0,y_0)# is the origin #(0,0)#, and the point in which it falls has coordinates: #(0,x)# (#x# is the range!), so:

#0=0+sinalpha*(x-0)/cosalpha-1/2g(x-0)^2/(v_0^2cos^2alpha)rArr#

#x*sinalpha/cosalpha-g/(2v_0^2cos^2alpha)x^2=0rArr#

#x(sinalpha/cosalpha-g/(2v_0^2cos^2alpha)x)=0#

#x=0# is one solution (the initial point!)

#x=(2sinalphacosalphav_0^2)/g=(v_0^2sin2alpha)/g#

(using the double-angle formula of sinus).

Now let's find, finally #alpha#.

#sin2alpha=(gx)/v_0^2=(9.8*120)/250^2=0.0188°rArr#

#2alpha_1=arcsin0.0188rArr2alpha=1,078°rArralpha=0.53°# and

#2alpha_2=180°-arcsin0.0188=178.92°rArralpha=89.47°#.

The solution are both valid, but the most reasonable is the first.

Using trigonometry, now, we can find the answer.
There is a right-angled triangle with an acute angle of #0.53°#, and a cathetus of #120m#. The other cathetus is

#c=120*tan0.53=1.11m#.