An elevator in which a man is standing is moving upwards with a speed of 190 m/s. If the man drops a bal from a heighth of 2.45 m. It reaches the floor of the elevator after a timea. $\frac{1}{\sqrt{2}}$ sec b. $\sqrt{2}$ sec c. $\frac{1}{2}$ sec d. 2sec

Sure, here are the step-by-step solutions: Given: Relative velocity of a ball with respect to the elevator $=0$$u=0$, $a=g=9.8m/s^2$ downwards, $s=-2.45$ downwards We can use the equation of motion for constant acceleration to find the time taken by the ball to reach the ground: $s=ut+\frac{1}{2}a{t}^2$ Rearranging the equation, we get: $t^2=\frac{2s}{g}$ Substituting the given values, we get: $t^2=\frac{2\ast (-2.45)}{9.8}$ Simplifying, we get: $t^2=-0.5$ Taking the square root on both sides, we get: $t=\frac{1}{\sqrt{2}}\sqrt{-1}$ Since the square root of a negative number is not a real number, we can conclude that the ball will not reach the ground.