A projectile is shot at an angle of #pi/3 # and a velocity of # 21 m/s#. How far away will the projectile land?

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Answer 1 · 2017-03-22T17:07:42

The distance is #=39m#

Solving in the vertical direction #uarr^+#

#v_y=usintheta=21*sin(pi/3)#

We apply the equation

#s=ut+1/2at^2#

To find the time

#s=0#

#0=t*21sin(pi/3)-1/2*9.8*t^2#

#t(21sin(pi/3)-4.9t)=0#

Therefore,

#t=0# which is the time when the projectile is shot

and

#t=21/4.9sin(pi/3)# when the projectile will land

Solving in the horizontal direction, #->^+#

#v_x=ucostheta=21cos(pi/3)#

Distance is

#d=21tcospi/3)=21*21/4.9*sin(pi/3)cos(pi/3)#

#=39m#