A ball A is projected with speed 50 ms^-1 at an angle 37^o with horizontal. At the same time, a second ball B is projected in a direction shown in the figure with speed 60 ms^-1 from a point at a distance of 91 m from A. Both ball collide in air. (Given, $\sqrt{3}$ = 1.7). The value of $θ$ (in degrees) is

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Detailed Solution

Since, both balls are at same levels, so for collision, $V_{AY} = V_{BY}$

$\begin{array}{l} or u_{AY} - gt = u_{BY} - gt \\ \Rightarrow u_{AY} = u_{BY} \\ \Rightarrow 50 \sin 37^{\circ} = 60 \sin θ \\ \Rightarrow \sin θ = \frac{50 \times \frac{3}{5}}{60} \\ \Rightarrow θ = 30^{\circ} \end{array}$

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