Question

An elevator at rest which is at 10th floor of a building is having a plane mirror fixed to its floor. A
particle is projected with a speed V2 m/s and at 45° with the horizontal as shown in the figure. At the
very instant of projection, the cable of the elevator breaks and the elevator starts falling freely. What
will be the separation (in m) between the particle and it's image 0.5 s after the instant of projection?​

Answer 1

Answer:

For particle:

h=1×0.5−

2

1

×10×

4

1

=

2

1

−

4

5

=−

4

3

for elevator

h

1

=−

2

1

×10×

4

1

=−

4

5

then distance between particle and mirror

=

4

5

−(−

4

3

)=−

4

1

then if

u=

4

1

=

4

1

+

4

1

=

4

2

=

2

1

m

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