In the given figure point A and C are on the horizontal ground & A and B are in same vertical plane. Simultaneously bullets are fired from A, B and C and they collide at D. The bullet at B is fired horizontally with speed of $\frac{72}{5}$ km/hr and the bullet at C is projected vertically upward at velocity of $\frac{54}{5}$ km/hr. Find velocity of the bullet projected from A in m/s.

Velocity of bullet B, $V_B=\frac{72}{5}km/h=4m/s$
Velocity of bullet C, $V_C=\frac{54}{5}km/h=3m/s$
In this situation, for the three bullets to collide
Along x-axis
Horizontal distance covered by the bullet A must be equal to the horizontal distance covered by bullet B
$\Rightarrow u\cos \theta \times t=4\times t\Rightarrow u\cos \theta =4m/s$
Along y-axis
Vertical distance covered by the bullet A must be equal to the vertical distance covered by bullet C
$\Rightarrow u\sin t-\frac{1}{2}\times g{t}^2=3\times t-\frac{1}{2}g{t}^2\Rightarrow u\sin \theta \times t=3\times t\Rightarrow u\sin \theta =3m/s$
Hence, velocity of projection of bullet from $A=\sqrt{3^2+4^2}=5m/s$
Alternatively:
For collision $u=\sqrt{u_c^2+u_B^2}=\sqrt{{(\frac{72}{5}\times \frac{5}{18})}^2+{(\frac{54}{5}\times \frac{5}{18})}^2}=\sqrt{4^2+3^2}=5m/s$