Question

From a tower of height 19.6 m two bodies are simultaneously projected horizontally in opposite directions, with velocities of 1 m/s and 4 m/s, respectively.
The horizontal distance between the two bodies when their velocity vectors are perpendicular to each other is

Moderate

Solution

Time after which velocities become perpendicular to each other $T = \frac{\sqrt{u_{1} u_{2}}}{g}$
The horizontal distance between the bodies at the instant when the velocity vectors are mutually perpendicular, we have $x = v_{1} t + v_{2} t = (5) (0 .2) = 1 m$

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