A projectile is shot at an angle of #pi/4 # and a velocity of # 21 m/s#. How far away will the projectile land?

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Answer 1 · 2017-04-09T06:50:46

The distance is #=45.1m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=21*sin(1/4pi)#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=21sin(1/4pi)-g*t#

#t=21/g*sin(1/4pi)#

#=1.52s#

Resolving in the horizontal direction #rarr^+#

To find the distance where the projectile will land, we apply the equation of motion

#s=u_x*2t#

#=21cos(1/4pi)*1.52*2#

#=45.1m#