Question

A boat is moving directly away from the gun on the shore with speed v₁. The gun fires a shell with speed v₂ at an angle α and hits the boat. The distance of the boat from the gun at the moment it is fired is

Moderate

Solution

See fig. (7)
For shell to hit the boat, the time taken by the boat to go from B to C with velocity v₁ should be equal to the time taken by the shell to go from O to A and then A to C, i.e.,
$\frac{2 v_{2} \sin α}{g} = T$
Further, $x + v_{1} T = R = \frac{2}{g} (v_{2} \sin α) (v_{2} \cos α)$
$\begin{array}{l} or x + \frac{2}{g} v_{1} v_{2} \sin α = \frac{2}{g} (v_{2} \sin α) (v_{2} \cos α) \\ or x = \frac{2}{g} (v_{2} \sin α) (v_{2} \cos α) - \frac{2}{g} v_{1} v_{2} \sin α \\ or x = \frac{2}{g} (v_{2} \sin α) [(v_{2} \cos α) - v_{1}] \end{array}$

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