A projectile is shot from the ground at an angle of #pi/4 # and a speed of #6 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2016-04-17T12:22:05

#2.6m#

Split up the velocity into its horizontal and vertical components using trigonometry,

#V_h=6m/s*cos(pi/4)=6cos45#
#V_v=6m/s*sin(pi/4)=6sin45#

Now you should work out the halfway point, where upwards velocity becomes #0# and the projectile reaches its maximum point

#v=u+at#
#0=V_v+"gt"#,

where the acceleration of gravity #g=-9.8#

#0=6sin45-9.8t#

#t=(6sin45)/9.8#
#t=0.433s#

Now you can work out the horizontal distance by

#s=vt#
#s=V_h*0.433#
#=6cos45*0.433=1.84m#

and the vertical distance by

#s=V_v*0.433#
#=6sin45*0.433=1.84m#.

The overall distance is found by Pythagoras' theorem,

#c=sqrt(a^2+b^2)#

#c=sqrt(1.84^2+1.84^2)=2.6m#