A projectile is shot at an angle of #pi/6 # and a velocity of # 48 m/s#. How far away will the projectile land?
2 Answers
Explanation:
Range of the projectile is given by the formula
#"R" = ("u"^2sin2theta)/"g"#
The range of the projectile is about
Explanation:
We can find the projectile's total horizontal displacement using kinematics. We will first make the assumption that air resistance and all other outside forces are negligible and consider only the effects of the force of gravity.
We have the following information:

#theta=pi/6# 
#v_i=48" m"//"s"# 
#a_y=g=9.81" m"//"s"^2# 
#a_x=0" m"//"s"^2#
My approach would be to calculate the time elapsed over the motion using the vertical components and use this to find the horizontal displacement.
We will use the following kinematic equation and solve for
#color(blue)((v_f)_y=(v_i)_y+a_yDeltat),#
using the fact that at the top of the object's motion,
Before we can use this equation, however, we will have to find
 This diagram represents only one velocity vectorthat is, the first (
#v_i# ).
#sin(theta)="opposite"/"hypotenuse"#
#=>sin(theta)=v_y/v#
#=>color(blue)(v_y=vsin(theta))#
#=>=48" m"//"s"*sin(pi/6)#
#=>=(48)/2#
#=>=color(blue)(24" m"//"s")#
We can now make use of the above kinematic equation and solve for the flight time from laugh to maximum altitude.
#(v_f)_y=(v_i)_y+a_yDeltat#
#=>Deltat=(cancel(v_f)v_i)/a#
#=(24" m"//"s")/(9.81" m"//"s"^2)#
#~~color(blue)(2.45"s")#
Assuming the launch and landing altitudes are equal (as we are not told otherwise), the total flight time is twice the rise time, or
We can now put this time into a new equation to find
#Deltax=(v_i)_xDeltat+cancel(1/2a_xDeltat^2)#
#Deltax=(v_i)_xDeltat#
As above, we find the parallel component of the initial velocity using trigonometry. We get
#=(48"m"//"s")cos(pi/6)(4.9" s")#
#~~203.7"m"#
#=>color(blue)(204"m")#