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?

Question

1208 views around the world

You can reuse this answer
Creative Commons License

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#