Suppose that you launch a projectile at a high enough velocity that it can hit a target at a distance. Given the velocity is 34-m/s and the range distance is 73-m, what are two possible angles the projectile could be launched from?

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Answer 1 · 2015-06-12T15:21:55

#alpha_1~=19,12°#

#alpha_2~=70.88°#.

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 we have the right formula to answer the question:

#sin2alpha=(x*g)/v_0^2=(73*9.8)/34^2~=0,6189rArr#

#2alpha_1~=arcsin0,6189+k360°~=38,23°#

#alpha_1~=19,12°#

and (the sinus has supplementary solutions):

#2alpha_2~=180°-arcsin0,6189+k360°~=180°-38,23°~=141,77°#

#alpha_2~=70.88°#.