A truck starts from rest and accelerates uniformly at 2.0 ms^-2.. At t = 10s, a stone is dropped by a person standing on the top of the trunk (6m high from the ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? (Neglect air resistance, Take g=10ms^-2)

$$\begin{gathered} 22.4 \mathrm{m}/\mathrm{s} \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{of} {\tan }^{-1}\frac{1}{2} \\ \mathrm{with} \mathrm{the} \mathrm{horizontal}. 10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{vertically} \mathrm{downwards} \end{gathered}$$

$$\begin{gathered} 22.4 \mathrm{m}/\mathrm{s} \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{of} {\tan }^{-1}\frac{1}{2} \\ \mathrm{with} \mathrm{the} \mathrm{vertical}. 10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{vertically} \mathrm{downwards} \end{gathered}$$

$$\begin{gathered} 10 \mathrm{m}/{\mathrm{s}}^2 \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{of} {\tan }^{-1}\frac{1}{2} \\ \mathrm{with} \mathrm{the} \mathrm{horizontal}. 22.4 m/s \\ \mathrm{vertically} \mathrm{downwards} \end{gathered}$$

$$\begin{gathered} 10 \mathrm{m}/{\mathrm{s}}^2 \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{of} {\tan }^{-1}\frac{1}{2} \\ \mathrm{with} \mathrm{the} \mathrm{horizontal}. 22.4 \mathrm{m}/\mathrm{s} \\ \mathrm{vertically} \mathrm{downwards} \end{gathered}$$