A projectile is shot from the ground at an angle of #pi/12 # and a speed of #7 /15 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 · 2017-03-27T10:25:02

The distance is #=0.005m#

Solving in the vertical direction #uarr^+#

#v_y=u_0sintheta=7/15sin(1/12pi)ms^-1#

#a=-g=-9.8ms^-2#

#v=0# at the maximum height

We apply the equation

#v=u+at#

#0=7/15sin(1/12pi) - g*t#

#t=(7/15sin(1/12pi))/g=0.012s#

This is the time to reach the maximum height

Solving in the horizontal direction #rarr^+#

#v_x=u_0costheta=7/15*cos(1/12pi)=0.451ms^-1#

Distance #=v_x*t=0.012*0.451=0.005m#