Two particles A and B are projected simultaneously from the two towers of height 10 m and 20 m respectively. Particle A is projected with an initial speed of $10\sqrt{2}$ m/s at an angle of ${45}^0$ with horizontal, while particle B is projected horizontally with speed 10 m/s towards A. If they collide in air, then the distance between the towers is

$$\begin{gathered} Let the two particles meet at a dis\tan ce y below the point A and at a horizontal dis\tan ce of x from point A. \\ {u}_{Ax}=u \cos \theta =10\sqrt{2}\cos {45}^o=10m/\sec \\ {u}_{Ay}=u \sin \theta =10\sqrt{2}\sin {45}^o=10m/\sec \\ {u}_{bx}=u \cos \theta =10m/\sec \\ {u}_{by}=u \sin \theta =0m/\sec \end{gathered}$$

$$\begin{gathered} The dis\tan ce travelled by A in vertically downward direction in time t is (10- y) \\ \therefore (10-Y)=(-10)\times t+\frac{1}{2}\times g{t}^2---\left(1\right) \\ The dis\tan ce travelled by B in vertically downward direction in time t is (20- y) \\ \therefore (20-Y)=\frac{1}{2}\times g{t}^2---\left(2\right) \\ solve equations \left(1\right) and \left(2\right) \\ time t=1 sec=particles A and B will meet after 1sec \\ x is the horizontal dis\tan ce travelled by A in this one sec, \\ {x}_A=u_{Ax}t=10\times 1=10m=u_{Ax}t=10\times 1=10m \\ then x_B is horizontal dis\tan ce travelled by B \\ {x}_B=u_{Bx}t=10\times 1=10m \\ horizontal dis\tan ce between 2 towers =x_A+x_B=10+10=20m \end{gathered}$$