Initial velocity,$u = 0, a = 2.0 \mathrm{m\ s^{-2}} , t = 10\ \mathrm{s}$ Let $v$ be the velocity of truck when the stone is dropped from it after t = 10 s. ∴ Using the relation, v = u + at, we get $v=0+2.0\times 10=20\ \mathrm{m}\ \mathrm{s}^{-1}$Horizontal velocity of the stone when it is dropped from the truck is $v_x = v = 20 \mathrm{m\ s^{-1}}.$As air resistance is neglected, so $v_x$ = constant.Motion in the vertical direction :Initial velocity of the stone, $v_y = 0$ at $t = 10\,\mathrm{s}$ acceleration, $a_y = g = 10\ \mathrm{m\ s^{-2}}$, time $t = 11 - 10 = 1$ sIf $v_y$ be velocity of the stone after 1 s of drop (i.e. at t = 11 s,) then$v_{y}=u_{y}+a_{y} t$$=0+10\times 1=10\ \mathrm{m}\ \mathrm{s}^{-1}$If v be the velocity of the stone after 11 s, then$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$$=\sqrt{(20)^{2}+(10)^{2}}$ $=\sqrt{500}=22.4\ \mathrm{m}\ \mathrm{s}^{-1}$Let q be the angle which the resultant velocity OC of the stone makes with the horizontal direction OA i.e. with $v_x$. Then from ΔOAC,$\tan \theta = \frac{AC}{OA} = \frac{v_y}{v_x} = \frac{10}{20}=0.5$$\therefore \theta = 26.6^{\circ}$(b) At the moment, the stone is dropped from the truck, the horizontal force on the stone is zero,so, $a_x = 0$ and $a_y =$ acceleration along vertical direction $= +g = 10\ \mathrm{m\ s^{-2}}$ which acts in downward direction.∴ If a = resultant acceleration of the stone, then$a=\sqrt{a_{x}^{2}+a_{y}^{2}}$$=\sqrt{0^{2}+(10)^{2}}$ or $a=10\ \mathrm{ms^{-2}}$and it acts vertically downward.