A ball is dropped from a height of 49 m. The wind is blowing horizontally. Due to wind a constant horizontal acceleration is provided to the ball. Choose the correct statement (s). $(Take g=9.8{ \mathrm{ms}}^{-2})$

Distance travelled in vertical direction =49 m

$\therefore 49=\frac{1}{2}\times 9.8\times {\mathrm{t}}^2 \Rightarrow \mathrm{t}=\sqrt{10}=3.16 \mathrm{s}$

Let horizontal acceleration be $a {\text{m/s}}^2$

$$\begin{gathered} Distanc{e}_{\text{horizontal }}=x=0.5{\mathrm{at}}^2 \\ Distanc{e}_{\text{vertical }}=y=0.5\times 9.8{\mathrm{t}}^2=4.9{\mathrm{t}}^2 \end{gathered}$$

Hence, $\frac{y}{x}=\text{constant}$

Both values of horizontal and vertical distances covered progress in constant proportion to each other. Hence the ball follow is straight line.

Also since there is vertical movement too, total distance travelled is more than 49 m.