A boy throws a water-filled balloon at an angle of ${53}^{\circ }$ with a speed of $10m/s$. A car is advancing towards the boy at a constant speed of $5m/s$. If the balloon is to hit the car, how far away should the car be when the balloon is thrown? $(g=10m{s}^{-2})$

The correct option is D $17.6m$
InitiallyFinally

Given $u=10m/s,\theta ={53}^{\circ }$ So,
Time of flight, $T=\frac{2u\sin \theta }{g}=\frac{2\times 10\times \frac{4}{5}}{10}=\frac{8}{5}s$
Let the point of projection and point at which balloon hits the car be at same level, then range$R=\frac{u^2\sin 2\theta }{g}=\frac{100}{10}\times 2\times \frac{4}{5}\times \frac{3}{5}=\frac{48}{5}=9.6m$
Le the car travels distance $x_c$ towards boy before balloon hits the car, then
$x_c=v_{c}T=5\times \frac{8}{5}=8m$
Car's initial position, $\Rightarrow R+x_c=9.6+8=17.6m$