Answer
14.05 m/s
Work Step by Step
Let's apply equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find its y-component of velocity at a height of -10 m.
$\downarrow V^{2}=u^{2}+2aS $ ; Let's plug known values into this equation.
$V_{y}^{2}=0^{2}+2(9.8\space m/s^{2})10\space m$
$V_{y}=14\space m/s$
The x-component of his velocity remains constant (1.2 m/s) throughout the motion.
By using the Pythagorean theorem, we can get.
$V=\sqrt {V_{x}^{2}+V_{y}^{2}}=\sqrt {(1.2\space m/s)^{2}+(14\space m/s)^{2}}=14.05\space m/s$
