A particle is projected at an angle of ${37}^{\circ }$ with an inclined plane as shown in figure. Calculate:

(i) Time of flight of particle.

(ii) Distance traveled by particle (AB) along the inclined plane

The correct option is A

Time of flight = $\frac{12}{5}$sec

Distance along incline = $\frac{12}{5}(8-6\sqrt{3})m$

(i) To find out time of flight here, we can analyze the motion in y-direction; we can use the formula $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$. By analyzing motion in y-direction , the displacement of the particle in y-direction during motion is zero.

Now $u_y=usin\alpha =u.sin{37}^{\circ }=\frac{3}{5}\times 10=6\frac{m}{s}$

$a_y=-gcos\theta =-gcos{60}^{\circ }=-10\times \frac{1}{2}=-5\frac{m}{s^2}$

So, $y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow 0=6t-\frac{5}{2}{t}^2\Rightarrow t=\frac{12}{5}s$

(ii) To find out the distance traveled along AB, we have to analyze the motion in x-direction. So we have to use the formula

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Here $u_x=ucos\alpha =10cos{37}^{\circ }=10\times \frac{4}{5}=8\frac{m}{s}{a}_x=-gsin\theta =-10sin{60}^{\circ }=-10\times \frac{\sqrt{3}}{2}=-5\sqrt{3}\frac{m}{s^2}$

And $t=\frac{12}{5}s,x=8\times \frac{12}{5}-\frac{1}{2}.5\sqrt{3}{\left(\frac{12}{5}\right)}^2=\frac{96}{5}-\frac{5\sqrt{3}}{2}\times \frac{144}{25}=\frac{96}{5}-\frac{72\sqrt{3}}{5}=\frac{12}{5}(8-6\sqrt{3})m$