The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2
. If it were launched at an
angle $\theta $0 with speed v0 then (g = 10 ms–2) :
${\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$ and ${v_0} = {5 \over 3}$ ms-1
${\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$ and ${v_0} = {3 \over 5}$ ms-1
${\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$ and ${v_0} = {3 \over 5}$ ms-1
${\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$ and ${v_0} = {5 \over 3}$ ms-1
